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?

An object is thrown in the upward direction. At the point when it reaches the maximum height, the vertical component of its velocity will become zero.

The time for which the ball was in the air, $t=3.4\mathrm{s}$.

Assume that the initial speed with which the ball was thrown is $v_{\circ }$, the maximum height achieved by the ball is h, and the acceleration due to gravity is $\mathrm{g}=-9.8{\mathrm{ms}}^{-2}$. (Taking the upward direction as positive)

From the second equation of motion, you can write the displacement of the ball as

$s=v_{\circ }\times t+\frac{1}{2}\mathrm{g}{t}^2$.

Here, as the initial and final positions are the same, the displacement of the ball is zero. Thus,

$0=v_{\circ }\times 3.4+\frac{1}{2}\times \left(-9.8\right)\times 3.4^2$.

Solving the above equation for$v_{\circ }$,

$v_{\circ }=16.66{\mathrm{ms}}^{-1}$.

Thus, the initial velocity of the stone is $16.66{\mathrm{ms}}^{-1}$. (Positive sign indicates upward direction)

When the ball reaches its maximum height, its velocity will become zero. Thus, using the third equation of motion, you can write

$0^2-v_{\circ }^2=2\mathrm{g}h$.

Substituting the values in the above equation,

$0-16.{66}^2=2\times \left(-9.8\right)\times h$.

Solving for the value of h,

$h=14.16\mathrm{m}$.

Thus, the maximum height achieved by the ball is 14.16 m.