Question

A ball is thrown straight up with a speed of 15m/s near the surface of the earth. What is the Maximum height of the ball?(neglect air friction)

Answer 1

We can find the maximum height of the ball from the kinematic equation:

$$v^2=v_0^2+2gh,$$

here, $v=0\ \dfrac{m}{s}$ is the speed of the ball at the maximum height, $v_0=15\ \dfrac{m}{s}$ is the initial speed of the ball, $h$ is the maximum height of the ball and $g=-9.8\ \dfrac{m}{s^2}$ is the acceleration due to gravity.

Then, from this equation we can find the maximum height of the ball:

$$h=\dfrac{-v_0^2}{2g}=\dfrac{-(15\ \dfrac{m}{s})^2}{2\cdot (-9.8\ \dfrac{m}{s^2})}=11.5\ m.$$

Answer:

$h=11.5\ m.$