One Dimensional motion

Question

A tennis player hits a ball straight up in the air at a velocity of 3m/s l. How long will it take for the ball to return to the height from which it was hit?

Answer 1

It's pretty bold to ignore air resistance with an object as fuzzy as a tennis ball, so take this as only an estimate. This time will be an overestimation:

The one dimensional (1D) trajectory formula, with h(t) being the elevation above the point of release:

h(t) = h(0) + v(0)t - 0.5gt²

One things about physical problem solving is you get to control how you solve the problems. In this case I choose a 1D coordinate system h(t) that is zero at the point of release and positive above that: h(t) ≥ 0 where t=time in seconds.

When I write h(t) that means h as a function of t not h times t.

Since we assumed h(0)=0 and were given v(0)=3 m/s

h(t)=3t-0.5gt² and with g=9.81 m/s²,

h(t)=3t-4.9t²=t(3-4.9t)

So the ball is at zero elevation twice:

A. When t=0 and once when t=3/4.9=0.61 sec, and

B. When 3-4.9t=0

From B., we have t=-3/-4.9=0.61 sec.

Solution: The ball is in the air no longer than 0.61 sec.

Answer 2

On the upward flight, the velocity v = -9.8t + vo where t = time and vo = initial velocity

When v = 0 at the top, then t = 3/9.8 = .306 seconds

The time will double to return to the spot where it was hit so

t = 2(.306) = .612 sec