Wrongly called for a foul, an angry basketball player throws the ball straight down to the floor. If the ball bounces straight up and returns to the floor 2.7 s after first striking it, what was the ball's greatest height above the floor?

Question

anonymous · Answer

what we come to know from th above info. is that the time of flight is 2.7s. So, can you tell me what time of flight is?

anonymous · Answer

@xkat ?

matt101 · Answer

The total time it takes for the ball to return to the ground after first striking it is 2.7 s. This means the time it takes for it to reach its maximum height, halfway through the trip, is 1.35 s, as is the time it takes to descend from this height. We know the velocity at the maximum height is 0 m/s and we know acceleration (-9.8 m/s^2). Using these three pieces of information, we need to find distance traveled.

$$\Delta x=v _{i}t+\frac12at^{2}$$

Since this equation requires vi and the only speed we know is 0 m/s at the max height, we'll use the part of the trip where the ball travels from this height back to the ground so that vi = 0. Plugging in all numbers you'll get x = -8.93 m. This means the ball travelled 8.93 m down, meaning it must have started 8.93 m up!