Mod - 4

Module-4: Motion in Two and Three Dimension A general solution for the two dimensional motion of an object thrown in a gravitational field is usually termed a projectile motion problem. A projectile is an object upon which the only force acting is gravity. There are a variety of examples of projectiles: (i) an object dropped from rest is a projectile; (ii) an object which is thrown vertically upwards is also a projectile; and (iii) an object is which thrown upwards at an angle is also a projectile. Therefore “a projectile is any object which once projected continues in motion by its own inertia and is influenced only by the downward force of gravity”. The path of a projectile is called the trajectory.

Given the original conditions with which the projectile is thrown we will proceed to find the components of the original velocity and then move on to answer the following questions: •

Path (trajectory) of the Projectile

•

How much time passes till the projectile is at the top of its flight?

•

How high does the projectile rise?

•

How much time passes till the projectile strikes the ground?

•

How far away does the projectile land from its starting point?

Original, or initial, conditions: The original conditions are the size of the velocity and the angle above the horizontal with which the projectile is thrown. :

vo

Original angle

:

θ

X component of original velocity

:

vox = vo cos (θ)

Y component of original velocity

:

voy = vo sin (θ)

θ x velocity V0x = V0cosθ

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y velocityV0y = V0sinθ

Original velocity V0

Original size of velocity

Path (trajectory) of the Projectile: The initial velocity of the projectile Vo at t = 0 makes an angle θ with the positive x direction. The x and y component of Vo is given by vox = vocos(θ) and voy = vosin(θ) (1) Because there is no horizontal component of acceleration and vertical component of velocity after any arbitrary time t is given by vx = vocos(θ)

and

vy = vosin(θ) +at

(2)

The magnitude of the resultant velocity vector, v =

(v

2 x

+ v y2

)

vy

(4)

vx

The displacement along x and y axis at any time t is given as x = vocos(θ)t and

After time t vy v

(3)

The angle θ that the velocity vector makes with the horizontal at that instant is given by,

tan θ =

y

vx

voy vo t=0 θ vox

x

Fig: The path of a projectile

(5)

y = vosin(θ)t +

From equation (5), t =

1 2 ayt 2

(6)

x v o cos θ

Substituting the value of t in equation (6) we get,

Therefore

y = v o sin θ

x 1 x2 + ay 2 v o cos θ 2 v o cos 2 θ

1 1 ∴ y = (tanθ )x + ( a y 2 )x 2 2 v o cos 2 θ

(7)

This is the equation of the projectile. Since θ, vo and ay = -g are constant, this equation has the form, y = ax + bx2, the equation of a parabola. Hence the trajectory of a projectile is a parabola.

How much time passes until the projectile is at the top of its trajectory? At the top of the trajectory the y, or upward, velocity of the projectile will be 0.0 m/s. The object is still moving at this moment, but its velocity is purely horizontal. At the top it is not moving up or down, only across. Notice that the object is still in motion at the top of the trajectory; however, its velocity is completely horizontal. It has stopped going up and is about to begin going down. Therefore, its y velocity is 0.0 m/s. We need to find out how much time passes from the time of the throw until the time when the y velocity of the projectile becomes 0.0 m/s. This y velocity at the top of the trajectory can be thought of as the final y velocity for the projectile for the portion of its flight that starts at the throw and ends at the top of the trajectory.

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We will call this amount of time 'the half time of flight', since the projectile will spend one half of its time of flight rising to the top of its trajectory. It will spend the second half of its time of flight moving downward. We can use the following kinematics equation:

Vf = V0 + at

y velocity = 0.0 m/s

Subscript it for y:

Vfy = V0 y + ay t Solve it for t:

t=

Vfy − V0 y ay

Plug in 0.0 m/s for Vfy :

t=

0.0m / s − V0 y

(8)

ay

If the original y velocity and the y acceleration, i. e., the acceleration due to gravity, are plugged into the above equation, it will solve for the amount of time that passes from the moment of release to the moment when the projectile is at the top of its flight. We know, voy = vosinθ

and

ay = - g

Then the time at which the vy is zero t =

− V 0 sin θ V 0 sin θ = −g g

How high does the projectile rise?

(9)

y displacement when at top

Here you need to find the displacement in the y direction at the time when the projectile is at the top of its flight. We have just found the time at which the projectile is at the top of its flight. If we plug this time into a kinematics formula that will return the displacement, then we will know how high above ground the projectile is at when it is at the top of its trajectory. Here is the displacement formula:

d = V0t +

1 2 at 2

We must think of this displacement in the y direction, so we will subscript this formula for y:

d y = V0 y t +

1 ay t 2 2

(10)

If now we plug in the half time of flight, which was found above, we will solve for the height of the trajectory, since the projectile is at its maximum height at this time.

voy = vosinθ, ay = -g,

t=

0.0m / s − V0 y ay

=

− V0 y −g

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Then d y = V 0 y

dy = Or

 V0 y 1 + (− g ) g 2  g

V0 y

   

2

1 2 1 2 Voy − Voy g 2g

1  2  1  −  = Voy  g 2g  1 2 = Voy 2g

∴dy =

1 2 v o sin 2 θ 2g

(11)

How much time passes until the projectile strikes the ground? With no air resistance, the projectile will spend an equal amount of time rising to the top of its projectile as it spends falling from the top to the ground. Since we have already found the half time of flight, we need only to double that value to get the total time of flight. Total time T = 2t We know t =

V 0 sin θ g

Then the total time T = 2

V 0 sin θ g

(12)

How far away does the projectile land from its starting point? The distance from the starting point on the ground to the landing point on the ground is called the range of the trajectory. This range is a displacement in the x direction. It is governed by the x velocity of the projectile. This x velocity does not change during the flight of the projectile. That is, whatever is the value of the x velocity at the start of the trajectory will be the value of the x velocity throughout the flight of the projectile. The x velocity remains constant because there are no accelerations in the x direction. The only acceleration is in the y direction, and this is due to the vertical pull of gravity. Gravity does not pull horizontally. Therefore, the calculation for the range is simplified. Let us start with the general displacement formula:

d = V0t +

Range

1 2 at 2

Since we are working in the x direction, we should subscript this equation for x:

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d x = V0 x t +

1 axt 2 2

Now, since the acceleration in the x direction is 0.0 m/s2, the second term in the above equation drops out, and we are left with:

d x = V0 x t The velocity in the x direction does not change. The projectile maintains its original x velocity throughout its entire flight. So, the original x velocity is the only x velocity the projectile will have. We could, therefore, think of the last equation as:

d x = Vx t

(13)

If we plug in the original x velocity for vx and the total time of flight for t, we will solve for the horizontal displacement, or range, of the trajectory.

Put vx = vocos(θ) Then, dx = vo cos θ Range dx = R =

t =2

− V0 y ay

=2

− v o sin θ −g

2v o sin θ v o2 2 sin θ cos θ = g g

v o2 sin 2θ g

(14)

When a projectile is projected some height h above the ground:

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Example After you have studied the projectile motion material at General Solution In Two Dimensions, No Air Resistance, you may want to to solve an example problem of your own creation. Make up an original velocity vector. Then see if you can find solutions to all the common questions: •

What is the half time of flight?

•

What is the total time of flight?

•

What is the height of the trajectory?

•

What is the range of the trajectory?

•

What is the vector displacement at a certain time?

•

What is the vector velocity at a certain time?

Consider an object thrown with a velocity 50 m/sec with an angle of 50 degree as shown in the following diagram:

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Finding the x-component of the original velocity: This diagram is useful for understanding the right triangle trigonometry that is used to find the x-component of the original velocity: The cosine function is used to find the x-component of the velocity. Since: cos(angle) = adjacent side / hypotenuse cos(angle) = vox / vo Solving for vx: vox = (vo)(cos(angle)) Plugging in your values for vo and the angle: vox = (50.00 m/s)(cos(50.00 degrees)) vox = (32.15 m/s) vox = 32.13 m/s

Finding the y-component of the original velocity: This diagram is useful for understanding the right triangle trigonometry that is used to find the y-component of the original velocity: The sine function is used to find the y-component of the velocity. Since: sin(angle) = opposite side / hypotenuse sin(angle) = voy / vo Solving for vy: voy = (vo)(sin(angle)) Plugging in your values for vo and the angle: voy = (50.00 m/s)(sin(50.00 degrees)) voy = (38.28 m/s) voy = 38.30 m/s

Calculation for half time of flight, or t1/2: This is the amount of time that passes from the throw till the projectile is at the top of the trajectory. In general: vf = vo + at In the y-direction: vfy = voy + ayt Then, doing some algebra: t = (vfy - voy) / ay And specifically for t1/2, since, (finally), at the top of the trajectory vfy = 0 m/s: t1/2 = (0 m/s - voy) / ay Plugging in your value for voy and -9.8 m/s2 for ay: t1/2 = (0 m/s - voy) / ay t1/2 = (0 m/s - 38.30 m/s) / -9.8 m/s2 t1/2 = 3.90 s

Calculation for the total time of flight, or tT: One way to get the total time of flight is to double the half time:

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tT = 2(t1/2) Using the above half time: tT = 2(3.90 s) tT = 7.81 s

Calculation for the maximum height of the trajectory: The maximum height is a displacement; so, for any displacement: d = vot + 0.5at2 The height is a vertical, or y, displacement; so, if we subscript the above equation for y, it looks like this: dy = voyt + 0.5ayt2 It takes half the time of flight for the projectile to get to the maximum height; so, we want to know the ydisplacement when the time equals t1/2. The above displacement equation with t1/2 entered as the time looks like this: dy = voyt1/2 + 0.5ayt1/22 Entering your values for voy and t1/2, and entering -9.8 m/s2 for ay we get: dy = (38.30 m/s)(3.90 s) + 0.5(-9.8 m/s2)(3.90 s)2 dy = (149.70 m) + (-74.85 m) dy = 74.85 m

Calculation for the range of the trajectory: The range is a displacement, and, again, for any displacement: d = vot + 0.5at2 The range is a horizontal, or x, displacement; so, if we subscript the above displacement equation for x, it looks like this: dx = voxt + 0.5axt2 The acceleration in the x-direction is zero, so this last equation changes to: dx = voxt + 0.5(0 m/s2)t2 This causes the last term to drop out; so, we are left with: dx = voxt We want this x-displacement for the total time of flight; so, finally, we get: dx = voxtT Now, plugging in your value for vox and tT we get: dx = (32.13 m/s)(7.81 s) dx = 251.22 m

Calculation for the displacement vector at the certain time: First, we must find the x- and y-components of the displacement vector at the certain time. Let's do the xcomponent first. As with the above range calculation, since there is no x-acceleration, the displacement xcomponent simplifies to: dx = voxt Plugging in your values for the original x-component of the velocity and the certain time we get: dx = (32.13 m/s)(1.00 s) dx = 32.13 m Next, we will get the y-component of the displacement vector. There is an acceleration in the y-direction; so, we will use the complete displacement equation, subscripted for y: dy = voyt + 0.5ayt2

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Now, plugging in -9.8 m/s2 for ay and plugging in your values for the original y-component of the velocity and the certain time we get: dy = (38.30 m/s)(1.00 s) + 0.5(-9.8 m/s2)(1.00 s)2 dy = (38.30 m) + (-4.90 m) dy = 33.40 m If you have selected a certain time less than the total time of flight, your displacement vector basically looks like this: And now we need to find the size of the two dimensional displacement vector, d, using the Pythagorean theorem: d2 = dx2 + dy2 d2 = (32.13 m)2 + (33.40 m)2 d2 = (1032.93 m2) + (1115.70 m2) d2 = 2148.64 m2 d = 46.35 m We will now use the arctangent function to find the angle for this displacement. We will consider the x- and y-displacements to form a simple right triangle with the two dimensional displacement vector, d, forming the hypotenuse. Without regard to the sign, (+ or -), of the components, but only treating this as a simple geometry problem with a right triangle, the size of the angle which the hypotenuse makes with the horizontal is: angle = arctan(|dy| / |dx|) This is shown in the following picture: Using your values for dy and dx: angle = arctan(33.40 / 32.13) angle = arctan(1.03) angle = 46.10 degrees (If you are using a calculator, be sure you are working with degrees and not radians.) If dy actually is pointing up, that is, if it is positive, then this angle is N of E. If dy is pointing down, being negative, then this angle is S of E. So, here is the vector displacement at the certain time, t = 1.00 s: d = 46.35 m angle = 46.10 degrees N of E

Calculation for the velocity vector at the certain time: First, we must find the x- and y-components of the velocity vector at the certain time. We will do the xcomponent first. In general: vf = vo + at So, for the x-component of the velocity vector: vfx = vox + axt Now, the acceleration in the x-direction, ax, is 0 m/s2, so the back term in the above equation drops out, as in: vfx = vox + (0 m/s2)t vfx = vox Basically, this demonstrates that the x-velocity does not change during the flight of the projectile. Using your value for the original x-velocity:

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vfx = 32.13 m/s Next, we will get the y-component of the velocity vector. There is an acceleration in the y-direction; so, we will use the complete velocity equation, subscripted for y: vfy = voy + ayt Now, plugging in -9.8 m/s2 for ay and plugging in your values for the original y-component of the velocity and the certain time we get: vfy = (38.30 m/s) + (-9.8 m/s2)(1.00 s)2 vfy = (38.30 m/s) + (-9.80 m/s) vfy = 28.50 m/s Perhaps your y-velocity is positive. If so, this would be your basic velocity diagram: Perhaps your y-velocity is negative. If so, this would be your basic velocity diagram: Next, we need to find the size of the two dimensional velocity vector, v, using the Pythagorean theorem: v2 = vx2 + vy2 v2 = (32.13 m/s)2 + (28.50 m/s)2 v2 = (1032.93 m2/s2) + (812.37 m2/s2) v2 = 1845.31 m2/s2 v = 42.95 m/s In a manner much like we worked with displacement angle, we will now use the arctangent function to find the velocity angle. We will consider the x- and y-velocities to form a simple right triangle with the two dimensional velocity vector, v, forming the hypotenuse. Again, without regard to the sign, (+ or -), of the components: angle = arctan(|vy| / |vx|) Using your values for vy and vx: angle = arctan(28.50 / 32.13) angle = arctan(0.88) angle = 41.56 degrees If vy actually is pointing up, that is, if it is positive, then this angle is N of E. If vy is pointing down, being negative, then this angle is S of E. A positive y-velocity makes the angle N of E: A negative y-velocity makes the angle S of E: Here is the vector velocity at the certain time, t = 1.00 s: v = 42.95 m/s angle = 41.56 degrees N of E

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Problem Sheet : Projectile Motion 1. A ball is projected at an angle 55o above the horizontal and with an initial velocity of 140 m/sec. (i) Find the position of the projectile and the magnitude and direction of its velocity when t = 1.5 sec. (ii) Find the time at which the projectile reaches the highest point of its flight, and find the elevation of this point. (iii) Find the horizontal range.

2. A placekicker kicks a football at an angle of 35o above the horizontal axis. The initial speed of the ball is 35 m/s. Ignore air resistance find (i) the maximum height that the ball attains, (ii) the time of flight between kickoff and landing and (iii) the range of the projectile.

3. A projectile is fired with an initial speed of 110m/s at an angle of 65 o above the horizontal from the top of a cliff 50 m high. Find (i) the time to reach the maximum height, (ii) the maximum height, (iii) the total time in the air, (iv) the horizontal range and (v) the components of the final velocity just before the projectile hits the ground. 4. A man stands on the roof of a building that is 32.0 m high and throws a rock with a velocity of magnitude 45.0 m/s at an angle of 35.0o above the horizontal. Air resistance been neglected. Calculate (i) maximum height above the roof by the rock, (ii) the magnitude of the velocity of the rock just before it strikes the ground (iii) the horizontal distance from the base of the building where the rock strikes the ground. 5. A football is thrown to a moving football player. The football leaves the quaterback's hands 1.6m above the ground with a speed of 14 m/s at an angle 30o above the horizontal. If the receiver starts 11 m away from the quarterback along the line of flight of the ball when it is thrown, what constant velocity must he have to get to the ball at the instant it is 1.6m above the ground? 6. A rescue plane is flying at a constant height of 1300 m with a speed of 450 km/h towards a point directly over a person struggling in water. At what angle of sight should the pilot release a rescue capsule if it is to strike very close to the person? 7. A plane drops a package of emergency rations to a stranded party of explorers. The plane is traveling horizontally at 42.0 m/s at 110 m above the ground. Find (i) where the package strikes the ground relative to the spot it was dropped and (ii) the velocity of the package just before it hits the ground. 8. A movie stuntman is to run across a rooftop and jump horizontally off it, to land on the roof of the next building (Fig.1). Before he attempts the jump, he wisely asks you to determine whether it is possible. Can he make the jump if his maximum rooftop speed is 4.2 m/s?

Man 4.8 m

6.2 m

9. A policeman chases a master jewel thief across city rooftops. They are both running at Fig: 1 5.5 m/s when they come to a gap between buildings that is 4.5 m wide and has a drop o of 3.5 m. The thief, having studied a little physics, leaps at 5.5 m/s and at 45 and clears the gap easily. The policeman did not study physics and thinks he should maximize his horizontal velocity, so he leaps at 5.5m/s horizontally. (i) Does he clear the gap? (ii) By how much does the thief clear the gap? 10. A pirate ship 450 m from a fort defending the harbor entrance of an island. The harbor defense cannon, located at sea level, has a muzzle velocity of 75 m/s.(i) To what angle must the cannon be elevated to hit the pirate ship? (ii) What are the times of flight for the two elevation angles calculated above? (iii) How far should the pirate ship be from the fort if it is to be beyond range of the cannon? 11. You toss a water balloon from your window 9.5 m above the ground. When the water balloon leaves your hand, it is moving at 12.0 m/s at an angle of 30o below the horizontal. How far horizontally from your window will the balloon hit the ground? Ignore air resistance.

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12. A rifle that shoots bullets at 420 m/s is to be aimed at a target 42.5 m away and level with the rifle. How high above the target must the rifle barrel be pointed so that the bullet hits target? 13. One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While you opponent is watching the first one, you throw a second snowball at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 25m/s. The first is thrown at an angle of 70o with respect to the horizontal. (ii) At what angle should the second snowball be thrown to arrive at the same point as the first? (ii) How many seconds later should the second snowball be thrown after the first to arrive at the same time?

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