A ball falls from a height of 2m and rebounds. If the mass of the ball is 60 g, then find the force between the ball and the ground. The time which the ball and ground was in contact is 0.2s. Please Answer Fast Urgent Urgent!!!
Answers
Answer:
Explanation:
Initial height = 10m
initial velocity = 0
velocity just before hotting ground = √2gh = √2*9.8*10 = 14.07 m/s (downward)
after rebound,
maximum height reached = 2.5m
final velocity at top = 0
initial velocity(just after rebound) = √2gh = √2*9.8*2.5 = √49 = 7 m/s (upward)
assuming downward as positive direction
So velocity just before hitting ground = +14.07 m/s
velocity just after hitting ground = -7 m/s
change in velocity = +14.07 - (-7) = 21.07 m/s
time = 0.01s
acceleration = change in velocity/time = 21.07/0.01
Answer:
While falling down, initial velocity = 0, final velocity with which the body strikes the ground = v m/s, acceleration due to gravity = 9.8 m/s 2 , height = 2m
The velocity of the ball with which it hits the ground can be found using,
v^ 2 = u 2 + 2as
=> v^ 2 = 0 + 2×9.8×2
=> v = 6.26 m/s (downward)
While it moves up, initial velocity = u, final velocity = 0, acceleration due to gravity = -9.8 m/s 2 , height = 0.5 m
The velocity of the ball with which it leaves the ground can be found using,
v^ 2 = u ^2 + 2as
=> 0 = u ^2 - 2×9.8×0.5 [this time the velocity and acceleration are oppositely directed so we have the negative sign]
=> u =3.13 m/s (upward)
So,Magnitude of the change in velocity = 6.26-3.13=3.13 m/s
=> Δv = 3.13m/s
∴ Impulse=change in linear momentum,dp = mΔv = 100 x 3.13=313 kg m/s
So, Force because of acceleration achieved during the strike is, F = dp/dt = 313/0.2 = 1565 N