Question

Two identical balls are shot upward one after another at an interval of $2s$ along the same vertical line with same initial velocity of $40m{s}^{-1}$ The height at which the balls collide is

• A. $50m$
• B. $75m$
• C. $100m$
• D. $125m$

Answer 1

The correct option is B $75m$

time taken for ball to rest is

$v=u+at$

$0=40-gt$

$t=4s$

therefore ball 1 is at rest at height $S=40\times 4-\frac{1}{2}\times 10\times 4^2=80m$

velocity of ball 2 after 2s as it was thrown after the delay of 2s

$v=u+at$

$v=40-10\times 2=20m/s$

$S_2=40\times 2-\frac{1}{2}\times 10\times 2^2=60m$

now ball 1 will drop with initial velocity of 0 and acceleration g and ball 2 with initial velocity of 20 m/s and deceleration of g and the separation is 20m

$S_1=0+\frac{1}{2}g{t}^2$

$S_2=20t-\frac{1}{2}g{t}^2$

$S_1+S_2=20$

$\frac{1}{2}g{t}^2+20t-\frac{1}{2}g{t}^2=20$

$t=1s$

$S_1=5m$

$S_2=15m$

hence ball 2 will be at 75m when they collide.