Question

The vertical height of $P$ above the ground is twice that of $Q$. A particle is projected downward with a speed of $9.8m/s$ from $P$ and simultaneously another particle is projected upward with the same speed of $9.8m/s$ from $Q$. Both particles reach the ground simultaneously. The time taken to reach the ground is (in $\sec$) ?

Answer 1

Answer: 6.00

For motion of particle thrown downwards from $P$,
$u=+9.8m/s,s=+2h$
$a=+9.8m/s^2,t=t$ (let)
$s=ut+\frac{1}{2}a{t}^2$
$2h=9.8t+\frac{1}{2}(9.8)t^2$ ...(1)
For motion of particle thrown upwards from $Q$,
$u=-9.8m/s,s=+h$
$a=+9.8m/s^2,t=t$
$s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow h=-9.8t+\frac{1}{2}(9.8)t^2$ ...(2)
Put value of $h$ from (2) to (1),
$2\left[-9.8t+\frac{1}{2}(9.8)t^2\right]=9.8t+\frac{1}{2}(9.8)t^2$
$-19.6t+(9.8)t^2=9.8t+4.9t^2$
$4.9t^2=29.4t$
$t=6\sec$