Question

A small ball mass 0.25kg thrown vertically upward reaches height of 18m . What is the minimum energy to reach this height?

Answer 1

V2−U2 = 2gS

Here given that, U = 0 m/s, S = 1.5 m. So,

V2−02 = 2×10×2 ⇒V=40−−√ m/s.

So, momentum of the ball is,

Pi=mV = 0.1×40−−√ = 0.1×6.325 = 0.633⇒ Pi=0.633 kg.m/s

After collision, it is given that the ball rises to a height S' = 1.5 m.

So, velocity with which the body rises is,

V2−U2=−2gS' ⇒ 02 − V'2 = −2×10×1.5 ⇒V' = 30−−√ m/s.

So, the momentum of the ball, when the ball is bouncing back.

Pf=mV' = 0.1×30−−√⇒Pf=0.547 kg m/s.

Therefore, the average force exerted by the floor on the ball is,

Favg=∆PDt = 0.633−0.54710−8=0.086×108 ⇒Favg=8.6×10^6N

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