Answer to Question #177795 in Physics for Safeer

Question

What is the velocity at the midway point of a ball able to reach a height y when thrown with upward with an initial velocity v0?

Answer 1

Let's first find the time that the ball takes to reach the maximum height:

$v=v_0-gt,$ $0=v_0-gt,$ $t=\dfrac{v_0}{g}.$

Then, we can find the maximum height reached by the ball:

$H=v_0t-\dfrac{1}{2}gt^2,$ $H=v_0\dfrac{v_0}{g}-\dfrac{1}{2}g(\dfrac{v_0}{g})^2=\dfrac{v_0^2}{2g}.$

Let's find the time that the ball takes to reach the midpoint:

$v=v_0-gt_1,$ $t_1=\dfrac{v_0-v}{g}.$

Then, we can find the midpoint height reached by the ball:

$h=v_0t_1-\dfrac{1}{2}gt_1^2,$ $h=v_0\dfrac{v_0-v}{g}-\dfrac{1}{2}g(\dfrac{v_0-v}{g})^2,$ $h=\dfrac{v_0^2-v^2}{2g}.$

Since $h=\dfrac{H}{2}$, we can write:

$\dfrac{v_0^2-v^2}{2g}=\dfrac{v_0^2}{4g},$ $2v_0^2-2v^2=v_0^2,$ $v=\dfrac{v_0}{\sqrt{2}}.$