A clown starts to juggle balls. For one act, he throws a ball vertically upwards and runs to a cage 6m away. He runs at a constant speed of 3 m/s and aims to touch the cage and return in time to...

A clown starts to juggle balls. For one act, he throws a ball vertically upwards and runs to a cage 6m away. He runs at a constant speed of 3 m/s and aims to touch the cage and return in time to catch the falling ball.

(i) Calculate the initial minimum speed with which the ball must be thrown upwards to accomplish this task.

(ii) How high does the ball reach?

Borys Shumyatskiy | Certified Educator

Hello!

It is obvious that the time needed for a clown to return is `t_0=4s` (twice 6m divided by 3m/s). A ball must return to the same height after the same time.

Choose a vertical upward axis starting at the initial height of a ball. Then the height of a ball is `H(t)=V_0 t-(g t^2)/2,` where `V_0` is the initial speed we have to find, `t` is time and `g` is the gravity acceleration.

We need `H(t_0)=0` (a ball returns at the same time a clown returns). It is a simple equation for `V_0,` `V_(0) t_(0)=(g t_0^2)/2,` or `V_0=(g t_0)/2.` In numbers it is `(9.8*4)/2=19.6 (m/s).` This is the answer for (i).

For (ii), use that speed of a ball is `V(t)=V_0 - g t,` and that at the maximum height the speed is zero. So the time when the maximum height is reached is `t_1=V_0/g,` and the corresponding height is `H(t_1)=(V_0^2)/(2g)=(g t_0^2)/8=(9.8*4^2)/8=19.6 (m).`

This is the answer for (ii).