A ball is dropped from a height 2.50 m above the floor, (a) Find the speed v with which it reaches the floor, (b) The ball now rebounds. The speed of the ball is decreased to 3u/4 due to this collision. How high will the ball rise?
Answers
=>e=v'/v
so v=√2×10×2.50=5√2m/s.
Also h'/h=e²..........(1)
h' is the height of separation and h=2.50m.
Then, H=h+e²h'
let,v=u Then, v'=u-3u/4=u/4=5/(2√2)m/s.
Again, e=1/4.
Then,using (1) we have ,
h'=(1/16)×2.5=0.15m.
==============ⓢⓦⓘⓖⓨ
==============ⓢⓦⓘⓖⓨ
Ball is dropped from a height, s = 90 m
Initial velocity of the ball, u = 0
Acceleration, a = g = 9.8 m/s2
Final velocity of the ball = v
From second equation of motion,
time (t) taken by the ball to hit the ground can be obtained as:
S = ut + 1/2 at^2
90 = 0 + 1/2 x 9.8 x t^2
t = √18.8 = 4.29 sec
From first equation of motion,
final velocity is given as:
v = u + at = 0 + 9.8 × 4.29 = 42.04 m/s
Rebound velocity of the ball,
u(r) =(9/10 )v= (9/10)× 42.04 = 37.84 m/s
Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:
v = ur + at′
0 = 37.84 + (– 9.8) t′
t' = -37.84/ -9.8 = 3.86 s
Total time taken by the ball = t + t′ = 4.29 + 3.86 = 8.15 s
As the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the floor for the second time. The velocity with which the ball rebounds from the floor
= (9/10)× 37.84= 34.05 m/s
Total time taken by the ball for second rebound = 8.15 + 3.86 = 12.01 s The speed-time graph of the ball is represented in the given attachment ( please see attached file)
I hope, this will help you
=======================
·.¸¸.·♩♪♫ ⓢⓦⓘⓖⓨ ♫♪♩·.¸¸.·
___________♦♦⭐♦ ♦___________