Question

A ball is kicked from the ground with a velocity of 38.50 ft/sec at an angle of 35 degrees. How many feet will the ball travel horizontally before it comes back to the ground on its first bounce?

Answer 1

Answer:

Horizontal distance is 142.12 feet.

Step-by-step explanation:

We have,

Velocity of the ball is 38.50 ft/sec. The angle of projection of the ball is 35 degrees.

It is required to find the horizontal distance covered by it before it comes back to the ground on its first bounce. It means we need to find the horizontal range of the projectile. Range of a projectile is given by :

$R=\dfrac{u^2\sin2\theta}{g}\\\\R=\dfrac{(38.5)^2\times \sin2(35)}{9.8}\\\\R=142.12\ ft$

So, the horizontal distance covered by it before it comes back to the ground on its first bounce is 142.12 feet.