A tennis ball is dropped on the floor from a height of 4m. It rebounds to a height of 3m. If the ball was in contact with floor for 0.01s, what was the average acceleration during contact?
Answers
Answer:
Average acceleration during contact is 1652.25 m/s^2
Explanation:
u = sqrt(2gH)
= -sqrt(2 × 9.8 × 4)
v = -sqrt(2gh)
= sqrt(2 × 9.8 × 3)
Average acceleration = (v - u) / t
= (v - (-u)) / t
= (v + u) / t
= [sqrt(2 × 9.8 × 3) + sqrt(2 × 9.8 × 4)] / 0.01
= 1652.25 m/s^2
Average acceleration during contact is 1652.25 m/s^2
Answer:
Aav = 1652.5
Explanation:
case .1 , when ball moves downward (drop on floor)
u1=0
s1= 4m
by 3rd eq.
v² =u²+ 2as
v1²=u1² + 2 gs taking g positive
v1² = 0 + 2(9.8)(4)
= sqrt[78.4]
Now,
Case:2
when ball goes upwards after rebouncing
u2=v1= sqrt[78.4]
s2 = 3m
again by 3rd eq
v2² = u2²+ 2-g s taking g negative
v2²={ sqrt[78.4] }²+ (-58.8)
for average acce.
Aav = V2 - V1 / delta t
= sqrt 78.4 - (- sqrt 19.9 ) / 0.01
= sqrt 78.4 + sqrt 19.9 / 0.01
Aav = 1652.25 m/s²