A ball of mass 400 gm is dropped from a height of 5m.a boy on the ground hits the ball vertically upwards with a bat with an average force of 100N so that it attains a vertical height of 20m.find the time for which the ball remains in contact with bat?
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u = sqrt(2gH)
= sqrt(2 × 10 m/s² × 5 m)
= 10 m/s
Ball hits the bat with 10 m/s velocity
v = sqrt(2gh)
= sqrt(2 × 10 m/s² × 20 m)
= 20 m/s
Velocity of ball just after hit is 20 m/s
F = ma
F = m (v - u) / t
t = m(v - u) / F
= 400 × 10⁻³ kg × (20 - 10) m/s / 100 N
= 0.04 seconds
Ball remains in contact with bat for 0.04 seconds
= sqrt(2 × 10 m/s² × 5 m)
= 10 m/s
Ball hits the bat with 10 m/s velocity
v = sqrt(2gh)
= sqrt(2 × 10 m/s² × 20 m)
= 20 m/s
Velocity of ball just after hit is 20 m/s
F = ma
F = m (v - u) / t
t = m(v - u) / F
= 400 × 10⁻³ kg × (20 - 10) m/s / 100 N
= 0.04 seconds
Ball remains in contact with bat for 0.04 seconds
pinkspink23:
the answer is 0.12
the answer is upposed to be 0.12
Oh! v and u are in opposites
*opposite direction
v - (-u) = v + u
I’ll edit my answer soon
yea
no dats k i got the answe by changing dat......tysm
You’re welcome
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