Question -
Answer -
Height from which theball is dropped = 90 m
The initial velocityof the ball, u = 0
Let v be the finalvelocity of the ball
Using the equation
v2 – u2 =2as ——–(1)
v12 –0 = 2 x 10 x 90
v1= 42.43m/s
Time taken for firstcollision can be given by the equation
v = u + at
42.43 = 0 + (10) t
t1 =4.24 s
The ball lossesone-tenth of the velocity at collision. So, the rebound velocity of the ball is
v2= v –(1/10)v
v2 =(9/10) v
v2= (9/10)(42.43)
= 38.19 m/s
Time taken to reachmaximum height after the first collision is
v = u + at
38.19 = 0 + (10)t2
t2 =3.819 s
Total time taken bythe ball to reach the maximum height is
T = t1 +t2
T = 4.24+ 3.819 = 8.05 s
Now the ball willtravel back to the ground in the same time as it took to reach the maximumheight = 3.819 s
Total time taken willbe, T = 4.24 + 3.819 + 3.819 = 11.86
Velocity after thesecond collision
v3 =(9/10) (38.19)
v3 =34.37 m/s
Using the aboveinformation speed- time graph can be plotted