Projectiles

Physics Reviewer Projectiles Kinematics in Two Dimensions Motions in two dimensions can be described in terms of the following variables:

Equations of Kinematics in Two Dimensions Variables

x Component

y Component

Displacement

x

y

Acceleration

ax

ay

Final Velocity

vx

vy

Initial Velocity

v0 x

v0 y

Elapsed Time

t

t

3.3 a/b

vx = v0x + axt

vy = v0y + ayt

3.4 a/b

x = ½ (v0x + vx) t

y = ½ (v0y + vy) t

3.5 a/b

x = v0xt + ½ axt2

y = v0y + ½ ayt2

3.6 a/b

vx2 = v x2 + 2axx

vy 2= v0y2 + 2ayy

0

The x part of the motion occurs exactly as it would if the y part did not occur at all. Similarly, the y part of the motion occurs exactly as it would if the x part of the motion did not exist. In other words, the x and y motions are independent of each other. As a result, each can be analyzed separately according to the procedures of onedimensional kinetics.

Projectile Motion

A general motion of objects moving through the air in two dimensions near the earth’s surface. An object projected horizontally will reach the ground at the same time as an object dropped vertically from the same height. Example 1. An airplane moving horizontally with a constant velocity of +115 m/s at an altitude of 1050 m releases a package that falls to the ground along a curved trajectory. Ignoring air resistance, determine the time required for the package to hit the ground. Reasoning: The time required for the package to hit the ground is the time it takes for the package to fall through a distance of 1050 m. In falling, it moves to the right as well as downward, but these two parts of the motion occur independently. Therefore, we can focus solely on the vertical part. We note that the package is moving initially in the horizontal or x direction, not in the y direction, so that, v0y = 0 m/s. Furthermore, when the package hits the ground, the y component of its displacement is y = -1050 m; a = -9.8 m/s2 y = v0y + ½ ayt2 -1050 m = 0 m/s + ½ (-9.8 m/s2) t2 t = 14.6 s Example 2. At the same time as the plane in Ex. 1 releases the package (A), a stationary balloon drops a package (B) which falls straight down to the ground from a height of 1050 m. Find the speed of package (A) and the direction of the velocity vector just before package (A) hits the ground. Reasoning: Since the speed of the package is given by v = √vx2 + vy2, it is necessary to know vx and v, just before the impact. The component vx is constant and has the value of 115 m/s. The component can be determined by using the data in Ex. 1. Thus, we expect the final speed of package (A) will be greater than 115 m/s. from vy = v0y+ at = 0 m/s + (-9.8 m/s2) (14.6 s) = - 143 m/s The speed of the package (A) just before impact is: v = [(115 m/s)2 + (-143 m/s)2]1/2 = 184 m/s

tan ө = -143/115 = -1.24; ө = -51.20 South of East 3. A diver runs horizontally with a speed of 1.2 m/s off a platform that is 10.0 m above the water. What is his speed just before striking the water? Given: vx= 1.2 m/s; y = - 10.0 m; a = - 9.8 m/s2 v = ? Solution:

[Eq. 2.9] v2 = v02 + 2ax = 2ax = 2(- 9.8 m/s2) (- 10.0 m) = 196 m2/s2 vy = 14 m/s v2 = (14)2 + (1.2)2 v = 14.1 m/s

4. A horizontal rifle is fired at a bull’s eye. The muzzle speed of the bullet is 670 m/s. The rifle is pointed directly at the center of the bull’s eye, but the bullet strikes the target 0.025 m below the center. What is the horizontal distance between the end of the rifle and the bull’s eye? Given: vx = 670 m/s; y = - 0.025 m; a = - 9.8 m/s2. Solution:

[Eq. 2.8]

R=?

y = v0yt + ½ at2 = ½ at2 2 t = 2 (- 0.025m)/- 9.8 m/s2 = 0.005102 s2 t = 0.07143 R = 670 m/s (0.07143 s) R = 47.86 m

Problems: 1. A placekicker kicks a football at an angle of Ө = 400 above the horizontal. The initial speed of the ball is 22 m/s. Ignore air resistance and find the maximum height the ball attains. Reasoning: The maximum height is a characteristic of the vertical part of the motion, which can be treated separately from the horizontal part. In preparation for making use of this fact, we calculate the vertical component of the initial velocity: v0y = v0sinӨ = + (22 m/s) sin 400 = +14 m/s fr.eq. 3.4y:

y = H = (vy2 – v0y2)/2ay = + 10 m

2. From the top of a cliff overlooking a lake, a person throws two stones. The stones

have identical initial speed, but stone 1 is thrown downward at an angle Ө below the horizontal, while stone 2 is thrown upward at the same angle above the horizontal. Neglect air resistance and decide which stone, if either, strikes the water with the greater velocity. Reasoning: We might guess that stone 1, being hurled downward, would strike the water with the greater velocity. This, however, is not the case because, at

some point in the path of stone 2, it returns to its initial height; here, the speed of stone 2 is v0, but due to projectile symmetry, its velocity is now directed at an angle Ө below the horizontal. At this point, stone 2 has a velocity identical to that of stone 1 when it is thrown downward off the cliff. From this point on, the velocity of stone 2 changes in exactly the same way as that for stone 1, so both stones strike the water with the same velocity.

Components of Velocity Neglecting air resistance, a ball thrown into the air at an angle will travel in a parabolic path. The velocity of the ball (V) has independent vertical (V) and horizontal (H) components; the horizontal component stays the same the entire time the ball is in the air, while the vertical component, the only component affected by gravity, changes continuously while the ball is aloft. © Microsoft Corporation. All Rights Reserved.

Gravitational Forces Because the Moon has significantly less mass than Earth, the weight of an object on the Moon’s surface is only onesixth the object’s weight on Earth’s surface. This graph shows how much an object that weighs w on Earth would weigh at different points between the Earth and Moon. Since the Earth and Moon pull in opposite directions, there is a point, about 346,000 km (215,000 mi) from Earth, where the opposite gravitational forces would cancel, and the object's weight would be zero. © Microsoft Corporation. All Rights Reserved.

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