Question

A golf ball thrown at an angle of 20 relative to the ground at a speed of 40 m/s. if the range of the motion is 55m. what is the maximum height that the ball reaches.

Answer 1

Answer:

Maximum height of the projectile is 5.00 m

Explanation:

As we know that the projectile range is the net horizontal distance moved by the object

Here we know that angle of projection is

$\theta = 20^o$

velocity of projection is 40 m/s

now we know the formula of range as

$R = \frac{v^2 sin2\theta}{g}$

so we have

$55 = \frac{40^2 sin(2\times 20)}{g}$

now we have

$g = 18.7 m/s^2$

Now maximum height of the projectile is given as

$H = \frac{v^2 sin^2\theta}{2g}$

$H = \frac{40^2sin^220}{2(18.7)}$

$H = 5 m$