Question

A ball player catches a ball 3.4 s after throwing it vertically upward. With what speed did he throw it, and what height did it reach?

Answer 1

I hope that u understand

Answer 2

Answer:

Speed of the ball when it was thrown = 16.67 m/s

Height reached by the ball = 14.16 s

Explanation:

Given time in the ball comes to the same height from where the ball was thrown is 3.4 s.

Let us assume that the ball was thrown with a speed of u in the upward direction. Since only gravitational force acts on the particle in the downward direction, it has a constant acceleration in the downward direction.

We also know that a particle has the same magnitude of velocity when it is under a free fall at a fixed height but direction is opposite i.e., initially in the upward direction and finally in the downward direction.

$\therefore \dfrac{-u-u}{3.4}= -9.8\\\Rightarrow \dfrac{2u}{3.4}= 9.8\\\Rightarrow 2u = 9.8\times 3.4\\\Rightarrow u = \dfrac{9.8\times 3.4}{2}\\\Rightarrow u = 16.67\ m/s$

Hence, the particle was thrown with a speed of 16.67 m/s.

Now, we also note that the particle has zero velocity at its maximum height.

So, from the initial position to the maximum height reached by the particle, we have

$0^2 = u^2+2as\\\Rightarrow s = \dfrac{v^2-u^2}{2a}\\\Rightarrow s = \dfrac{0^2-16.67^2}{2(-9.8)}\\\Rightarrow s =14.16\ m$

Hence, the maximum height reached by the ball is 14.16 m.