a tennis ball was held 4.5 ft above the surface of a tennis court and was thrown vertically downward with an initial velocity whose magnitude was 12 fps. determine the magnitude of the velocity of the ball just before it struck the surface of the court

Answer 1

This is engineering? Hmmmm.

Vf^2=Vi^2 + 2 ad

Vf^2=12f/s+2*-32f/s^2*-4.5

Vf^2=288

Vf=12 sqrt2 fps

check that

Answer 2

the square root of 288 comes out to 16.9 so that would be my answer right? how did you come up with 12 sqrt2.