Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 04

Years : DEC 2015

Coefficient of restitution (e) = 0.7

Before Rebound:-

$u = -10 m/s, s = -3, \\ a = -g = -9.81 m/s^2$

Using, $v = u + 2as \\ \therefore v^2=(-10)^2+2\times -9.81\times -3 \\ \therefore v^2=158.86 \\ \therefore v= 12.6040 m/s$

Ball strikes the ground with a velocity of 12.604 m/s

After Rebound:-

Velocity after first rebound $u = ev = 0.7 \times 12.604 = 8.8228 m/s$

At highest point, $v =0, a = -g = -9.81 m/s^2$ . Using,

$v_1^2=u_1^2=2as_1\\ \therefore 0^2 =(8.8228)^2+2\times -9.81 \times s_1 \\ \therefore 19.62s_1=77.8414 \\ \therefore s_1=3.9675 m$

Maximum height the ball can reach after hitting the floor = 3.9675 m.