Question

Two balls of equal masses are thrown upwards, along the same vertical direction at an interval of $2$ seconds, with the same initial velocity of $40m/s$. Then these collide at a height of (Take $g=10m/s^2$).

• A. $120m$
• B. $75m$
• C. $200m$
• D. $45m$

Answer 1

Answer: (B)

$75m$

Let the two balls collide $tsec$ after the first ball is thrown, and let $h$ be the height at which they collide.
For the first ball:

$h=40t-\frac{1}{2}g{t}^2$
And for the second ball:

$h=40(t-2)-\frac{1}{2}g(t-2{)}^2$
$40t-\frac{1}{2}g{t}^2=40(t-2)-\frac{1}{2}g(t-2{)}^2$
$2gt=80+20=100$
$t=5sec$

$\therefore h=40\times 5-\frac{1}{2}\times 10\times 5^2=75m$

Hence, option B is correct.