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

9.4 m

Step-by-step explanation:

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

`H = ((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 = ((20\sin40^\circ)^2)/(2*9.8) = (165.27)/(19.6) = 8.4\text{ m}`

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