i really need help with these two To test the quality of a tennis ball, you drop

Question

i really need help with these two

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.25 m. Suppose the ball is in contact with the floor for 15.0 ms. What is the magnitude of its average acceleration during the contact with the floor? A basketball player grabbing a rebound jumps 62 cm vertically. How much (total) time does the player spend in the top 20 cm of this jump? How much (total) time does the player spend in the bottom 20 cm of this jump?

Explanation / Answer

average acceleration=change in velocity/change in time

a)initial velocity just before hitting the floor=sqrt(2*9.8*4)=8.854 m/s

velocity with which it leaves the ground=sqrt(2*9.8*2.25)=6.64 m/s

as its direction is in opposite to that of initial velocity,

change in velocity=6.64+8.854=15.494 m/s

average acceleration=15.494/0.015=1032.93 m/s^2

b)time taken to cover 20 cm while going from bottom to top and and from top to bottom is same.

let that time be t.

then 0.5*9.8*t^2=0.2

t=0.202 sec

so total time=2*0.202=0.404 sec

b)in bottom 20 cm,time taken=0.1124 sec

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.