Sup

a rock is thrown vertically upward with a speed of 12m/s. exactly 1.00 second later, a ball is thrown up vertically along the same path with a speed of 20.0 m/s (a) at what time will they strike each other? (b) at what height will the collision occur? (c) Answer (a) and (b) assuming that the order is reversed: the ball is thrown 1.00 s before the rock. Let's generate an equation that expresses the position of each object in terms of time, where t = 0 corresponds to time at which the ball is thrown. Thus, for the position of the ball, we have yR = y0 + v0, R t + 12 at 2 . Similarly, for the position of the rock, we have yB = y0 + v0, B (t − 1) + 12 a(t − 1)2 . Make sure that it is clear to you why we are subtracting 1 from t in the equation for the position of the ball. Substituting in that y0 = 0m; v0, R = 12 ms ; v0, B = 20 ms ; a = −9.8 sm2 , we have the following equations: yR = (0m) + (12 ms )t + ( 12 )(−9.8 sm2 )t 2 and yB = (0m) + (20 ms )(t − 1) + ( 12 )( −9.8 sm2 )(t − 1) 2 . When the rock collides with the ball, it must be in the same place as the ball at the same time; that is, yR = yB . Hence, at the time of collision, we set the right-hand-sides of our two position equations equal to each other; hence, we have that 12t − 4.9t 2 = 20(t − 1) − 4.9(t − 1) 2 . Solve this equation however you’d like. You will end up finding that t = 1.3989 s . That’s part a. For part b, we want to find the height at which the collision occurs; to do this, simply substitute our value for the time of collision (1.3989 s) into either position equation. As it would turn out, the collision occurs at y = 7.1980 m . For part c, we’ll rewrite our equations, letting t = 0 correspond to the time at which the ball is thrown; using (t1) in place of t in the equation for the position of the rock, to indicate that it is thrown one second after the ball; and using t in place of (t-1) in the equation for the position of the ball, to indicate that it is thrown at t = 0. This gives us the equations yR = y0 + v0, R (t − 1) + 12 a(t − 1)2 and yB = y0 + v0, B t + 12 at 2 . By plugging in the values we listed earlier, we get that: yR = (0m) + (12 ms )(t − 1) + ( 12 )(−9.8 sm2 )(t − 1) 2 and yB = (0m) + (20 ms )t + ( 12 )(−9.8 sm2 )t 2 As before, we seek to find the point in time when yB = yR . So we set the right-hand sides of both position equations equal to each other and get: 12(t − 1) − 4.9(t − 1) 2 = 20t − 4.9t 2 . Solve this equation, and you’ll find that t = 9.3889 s . This number should immediately seem a little wonky. Indeed, when you plug in to find the height at which the collision occurs, you find that the collision occurs at -244.16327 meters. The physical interpretation of this is that if the ball and rock were to continue traveling for a long distance beneath the point from which they were thrust upwards, they would end up colliding. In actual fact, both would hit the ground long before they would hit each other (though I don’t recall if the problem specifically specified that the thrower was standing on the ground or anything). You may as well indicate that numerically, they would collide 244some meters below the point of release, while in actual fact, they’d probably strike the ground first.