Question

a ball of mass 50 gram 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 newton so that it attain a vertical height of 45 m the time for which the ball remain in contact with the bat is

Answer 1

hope this helps you....

Answer 2

Answer:

Time of contact, Δt = 1/80th of a second.

Explanation:

Given;-

    Mass, m = 50 gram

Initial Dropping Height, H₁ = 20m

Force from bat, F = 200 N

Final Rebounding Height, H₂ = 45m

Now;-

[Refer to the attached image for the diagram]

We know that for a body when dropped attains a velocity i.e. v becomes equal to $\sqrt{2gH}$. In this case, since there is no specific mention of the velocity we assume that the $\sqrt{2gH}$is the velocity in the case of both drop and rebound of the mass, m.

So, [Just Before Collision];

    Velocity, v = $\sqrt{2gH}$₁

                   v = √2 × (10) × 20

                   v = √20 × 20

                   v = 20 m/s

Initial Momentum, Pi = - m × v

                 Pi = - 20 × m  ____(1)

Also, [After the collision];

    Velocity, v = $\sqrt {2gH}$₂

                    v = √ 2 × 10 × 45

                    v = √ 200 × 45

                    v = √ 900

                   v = 30 m/s      

Final Momentum, Pf =  m × v          

                   Pf = 30 × m _____(2)

Now;-

∵        Force, F = ΔP/ Δt

              200 N = Pf - Pi / Δt

              200 N = 30×m - ( - 20×m)/ Δt

                     Δt = 50 × m/ 200

                     Δt = 50 × 50 / 200 × 1000  [since, mass, m= 50 gm]

                     Δt = 25/ 2000

                     Δt = 1/80 th of a second

Hence, the ball remain in contact with the bat for 1/80th of a second.

Hope it helps! ;-))