Question

A ball is dropped from a height 20m. A second ball is thrown from the same height after 1 second with an initial velocity 'u'. Both the balls hit the surface at the same time.
a.) Find the initial velocity of the second ball.
b.) Do both the balls hit the surface with same velocity? Explain your answer.(g=10m/s²)
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Answer 1

Step-by-step explanation:

Initial  energy Ei=PE+KE=mgh+21mv2

Energy when it reaches the ground: E=21mv′2 since PE is zero.

Given, E=21Ei

Since it rises to the  same height Ef=mgh+0 is the final energy

⇒21(mgh+21mv2)=Ef=mgh+0

v2=2gh=2×10×20=202

⇒v=20m/s

Answer 2

Answer:

1. The initial velocity of the second ball = 20m/s

b.) The balls hit the surface with same velocity = 2s

EXPLANATION ::

By the Given condition

Initial velocity of ball = u=0

Final velocity of ball= v (say)

Distance =20m

Acceleration a=10m/s^2

Let Time of fall = t

We know that

v ^2−u^2=2as

v^2−0=2×10×20=400

v=20m/s

Now v=u+at gives

20=0+10×t

t=2s