physics question

Question

As a tennis ball is struck, it departs from the racket horizontally with a speed of 28.0 m/s. The ball hits the court at a horizontal distance of 19.6 m from the racket. How far above the court is the tennis ball when it leaves the racket?

Answer 1

This is a standard Projectile motion problem where you have motion in the X & Y coordinate system.

Motion in the X, velocity is constant. Which means acceleration is 0

Motion in the Y, you have acceleration in the Y direction with a magnitude of g.

Given:

Vox = 28.0 m/s

X = 19.6 m

Find:

Yo = ?

Solution:

Motion in the X

X = Vox * t + Xo

Xo = 0

X = Vox * t

t = X/Vox = 19.6/28 = 0.7 seconds (Time of trajectory from the time that it starts to when it hits the ground)

Now we move on to motion in the Y direction

Y = (1/2) * ay * t^2 + Voy * t + Yo

Y = 0 (Vertical position when it hits the ground)

ay = -9.81 m/s^2

Voy = 0 (Initial velocity is in the horizontal only!)

0 = -0.5 * 9.81 * t^2 + Yo

Solution:

Yo = 0.5 * 9.81 * 0.7^2 = 2.40 m