Question

Two spherical balls having equal masses with radius of $5\text{cm}$ each are thrown upwards along the same vertical direction at an interval of $3\text{s}$ with the same initial velocity of $35\text{m/s}$, then these balls will collide at a height of ____ $\text{m}$.
(Take $g=10{\text{m/s}}^2$)

Answer 1

Step 1: Given data and formula used

Let $h$ be the height at which both balls collide.

Therefore, two balls will collide when :

$h_1=h_2=h$

Let the two balls collide $t\text{sec}$ after the first ball is thrown.

Using, $s=ut+\frac{1}{2}a{t}^2$,

Step 2: Calculation of The height at which balls collide is

$35t-\frac{1}{2}\times 10t^2=35(t-3)-\frac{1}{2}\times 10(t-3{)}^2$

$-5t^2=-105-5t^2-45+30t$

$30t=150$

$\therefore t=5\text{s}$

The height at which balls collide is,

$h=35t-\frac{1}{2}\times 10t^2$

$h=35\times 5-\frac{1}{2}\times 10\times 5^2$

$h=50\text{m}$

Correct answer $:50$