In last week's homework set you explored the bounce of a 0.50 kg basketball. Let's look at the basketball's upward path. Last week's problem set said that the basketball bounced back up from the floor vertically at a speed of 7.2 m/s. Let's assume that no one touches the ball as it bounces upward, reaches some maximum height, and falls back down to the floor (to bounce again). (a) How much time passes before the ball hits the floor again? (b) What is the maximum height above the floor the ball reaches?
In last week's homework set you explored the bounce of a 0.50 kg basketball. Let's look at the basketball's upward path. Last week's problem set said that the basketball bounced back up from the floor vertically at a speed of 7.2 m/s. Let's assume that no one touches the ball as it bounces upward, reaches some maximum height, and falls back down to the floor (to bounce again). (a) How much time passes before the ball hits the floor again? (b) What is the maximum height above the floor the ball reaches?
College Physics
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Transcribed Image Text:4.2. In last week's homework set you explored the bounce of a 0.50 kg basketball. Let's look at the
basketball's upward path. Last week's problem set said that the basketball bounced back up
from the floor vertically at a speed of 7.2 m/s. Let's assume that no one touches the ball as it
bounces upward, reaches some maximum height, and falls back down to the floor (to bounce
again).
(a) How much time passes before the ball hits the floor again?
(b) What is the maximum height above the floor the ball reaches?
Expert Solution
Theory and part a
This is a concept of vertical motion.
Here the basketball is launched at 7.2m/s from the ground.
The acceleration acting is due to gravity in downward direction.
H = V0t -1/2gt2 = 0m as ball returns to ground.
Thus, t = 2V0/g = 2×7.2/9.81
t = 1.468 seconds
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