Answer to Question #269673 in Mechanics | Relativity for Klara

Question

A soccer player on a horizontal playground kicks a ball away with a velocity

v0 under an angle α with the playground. The ball can be considered to be a projectile. Air resistance can be neglected and the gravitational acceleration is g. During the trajectory, the ball reaches a maximum height H.Now he kicks a second ball under the same angle but with double velocity. What is the maximum height the second ball will reach?

Answer 1

$\text{for first ball:}$

$v_y = v_0\sin\alpha$

$h_{max}= \frac{v_y^2}{2g}= \frac{(v_0\sin \alpha)^2}{2g}=H$

$\text{for second ball:}$

$v'_0= 2v_0$

$v'_y = v'_0\sin\alpha= 2v_0\sin\alpha$

$h'_{max}= \frac{(v'_0\sin \alpha)^2}{2g}= \frac{(2v_0\sin \alpha)^2}{2g}=4H$

$\text{Answer: }4H$