Question

A particle P is let fall vertically from a height to the ground. Another particle Q is projected vertically down 3s later with a velocity of 3m/s^s from the same height. If both particles reached the ground simultaneously. Calculate:

a. time by each particle to reach the ground

b. distance travelled by each particle to reach the ground.

Answer 1

Let time taken by first particle be $t.$

If second particle is projected with velocity$=3\frac{m}{sec}$

For first particle, $d=\frac{1}{2}gt^2$

For second particle, $d=3(t-3)+\frac{1}{2}g(t-3)^2$

Equating both the equations, we get

$\frac{1}{2}gt^2=3(t-3)+\frac{1}{2}g(t-3)^2$

$t=\frac{4}{3}sec$

Putting $t=\frac{4}{3}sec$ in any of the equations, we get $d=\frac{80}{9}m$