A body of mass 50 g is dropped from a height of 20 m. A boy on the ground hits the ball vertically upwards with a bat with an average force of 200 N, so that it attains a vertical height of 45 m. The time for which the ball remains in contact with the bat is (g = 10 m/s^2)
v^2 = u^2 + 2gh
h = -ut +1/2 at^2
The Attempt at a Solution
v = √2gh = √2*10*20 = 20 m/s
This will become 'u' for the second part.
h = 45 m
a = F/m = 200 / 50 / 1000 = 4000 m/s^2
I substituted u, h & a in h = -ut + 1/2 at^2
I didn't get the answer which is 1/80th of a second.
I have tried using it by impulse method also.
Like, impulse = change in linear momentum = f * dt
mv - mu = f * dt
u = 20 m/s
I calculated v by v = √u^2+2gh = 10 √13
I still didn't get the answer.
I cannot understand where I am going wrong.