A ball falls freely from a height of 45 m . When the ball is at a height of 25 m , it explodes into two equal pieces. One of them acquires an additional horizontal component of velocity equal to 10 m s - 1 , while its vertical component remains the same. The distance between the two pieces when…

Question

A ball falls freely from a height of 45 m . When the ball is at a height of 25 m , it explodes into two equal pieces. One of them acquires an additional horizontal component of velocity equal to 10 m s - 1 , while its vertical component remains the same. The distance between the two pieces when both strike the ground is

Aicharya · Accepted Answer

Let at the time explosion velocity of one piece of mass is (10

\hat{i}

). If the velocity of other be

{\vec{v}}_{2}

, then from conservation law of momentum (since there is no force in the horizontal direction), the horizontal component of

{\vec{v}}_{2}

, must be

- 10

\hat{i}

.

∴

The relative velocity of two parts in the horizontal direction

= 20 m s^{- 1}

Time taken by the ball to fall through 45 m,

= t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 45}{10}} = 3 s

and time taken by the ball to fall through first 20m,

t^{'} = \sqrt{\frac{2 h'}{g}} = \sqrt{\frac{2 \times 20}{10}} = 2 s

. Hence time taken by ball pieces to fall from 25 m height to ground

= t - t^{'} = 3 - 2 = 1 s

.

∴

The horizontal distance between the two pieces at the time of striking on the ground

= 20 \times 1 = 20

m

A ball falls freely from a height of 45 m . When the ball is at a height of 25 m , it explodes into two…

Solution