A boy wants to throw a can straight up and then hit it with second can. He wants

Question

A boy wants to throw a can straight up and then hit it with second can. He wants the collision to occur 4.0 m above the throwing point. In addition, he knows that the time he needs between throws is 3.0 s. Assuming he throws both cans with the same speed, what must the initial speed be? Please show all work, an answer is not enough.
A boy wants to throw a can straight up and then hit it with second can. He wants the collision to occur 4.0 m above the throwing point. In addition, he knows that the time he needs between throws is 3.0 s. Assuming he throws both cans with the same speed, what must the initial speed be? Please show all work, an answer is not enough.

Explanation / Answer

Hi, Let the two cans collide after a time 't' after throwing the second can. So total time the first can is in flight before collission = (t + 3) sec. Given both collided at a height of s = 4m. also given both cans were thrown with same initial speed = U (say).    From s = ut + (1/2)at^2 we have, for first can      4 = U * (t+3) - (0.5 * 9.8 * (t+3)^2)   for second can 4 = U * t - (0.5 * 9.8 * t^2) (since acceleration is the acceleration due to gravity which is downwards and the cans thrown upwards, a = -g = -9.8). solving the above two equations we get t = 0.25sec and U = 17.225 m/s    Hope this helps you.    Hope this helps you.

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