Ncert Exercises & Solutions

A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed of 40 m/s. Calculate the relative speed of the balls as a function of time.

Hint: Here relative acceleration of balls with respect to each other is zero.

Step 1: Calculate the velocities of both balls after time t.

For the ball dropped from the building, $u_{1} = 0, u_{2} = 40 m / s$
The velocity of the dropped ball after time t,

$\begin{array}{l} v_{1} = u_{1} + gt \\ v_{1} = gt \end{array} \begin{array}{l} (downward) \end{array}$

For the ball thrown up, $u_{2}$ = 40 m/s
The velocity of the ball after time t

$\begin{array}{l} v_{2} = u_{2} - gt \\ = (40 - gt) \end{array} \begin{array}{l} (upward) \end{array}$

Step 2: Calculate the relative velocity

$∴$ The relative velocity of one ball w.r.t. another ball

$\begin{array}{l} = v_{1} - v_{2} \\ = gt - [- (40 - gt)] = 40 m / s . \end{array}$