Question

A ball is thrown from 1 m above the ground. The initial velocity is 20 m/s at an angle of 40 degrees above the horizontal. What is the maximum height of the ball above the ground

Answer 1

Answer:

9.4 m

Explanation:

This is a projectile motion. The maximum height above the level projection is given by

$H = \dfrac{(v_0\sin\theta)^2}{2g}$

$v_0$ is the initial velocity, $\theta$ is the angle of projection above the horizontal and $g$ is the acceleration of gravity.

$H = \dfrac{(20\sin40^\circ)^2}{2\times9.8} = \dfrac{165.27}{19.6} = 8.4\text{ m}$

The maximujm height above the ground is 1 m + 8.4 m = 9.4 m