f.4 physics

2007-06-17 1:21 am

A ball is released from a height of 5 m above the ground and it moves up to 2 m after rebound. What is the ratio of the speed of the ball just before hitting the ground to that just after rebounding from the ground?

a. 2.50:1
b 1.67:1
c. 1.58:1
d. 1.50:1

回答 (1)

Gabriella Montez

2007-06-17 1:32 am

✔ 最佳答案

Take downwards as +VE.
The speed before rebound (v1) :
v1^2 - u1^2 = 2as
v1^2 - 0^2 = 2(10)(5)
v1 = 10m/s

The speed aftr rebound (v2) :
0 - v2^2 = 2as'
0 - v2^2 = 2(10)(-2)
v2^2 = 40
v2 = 6.32m / s

v1 : v2
= 10 : 6.32
= 1.58 : 1 ( cor. to 3 s.f.)

Ans : C

2007-06-16 17:37:29 補充：
When the ball hits the ground, some of the emergy is converted into heat energy, sound energy, etc. The remaining energy is converted into PE for the rebound. At the height of 2m, all the KE has changed to PE and so at that point, the velocity is 0m/s.

參考: My Physics Knowledge

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