Asked by Jen
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
Answers
Answered by
bobpursley
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
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
Answered by
Juan
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.
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