Question

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

Answer 1

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