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(C) Let the height of point $Q$ be $h$ and the height of point $P$ be $2h$. Let the time taken to reach the ground be $t$. Taking the downward directi

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Both $(a)$ and $(b)$ are correct

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(C) Let the height of point $Q$ be $h$ and the height of point $P$ be $2h$. Let the time taken to reach the ground be $t$. Taking the downward direction as negative and the ground as the origin $(y=0)$:For particle at $P$: Initial velocity $u_P = -5 \, m/s$,displacement $s_P = -2h$. Using $s = ut + \frac{1}{2}at^2$,we get $-2h = -5t - 5t^2$,which simplifies to $2h = 5t + 5t^2$ (Equation $1$).For particle at $Q$: Initial velocity $u_Q = +5 \, m/s$,displacement $s_Q = -h$. Using $s = ut + \frac{1}{2}at^2$,we get $-h = 5t - 5t^2$,which simplifies to $h = 5t^2 - 5t$ (Equation $2$).Substitute $h$ from Equation $2$ into Equation $1$: $2(5t^2 - 5t) = 5t + 5t^2 \Rightarrow 10t^2 - 10t = 5t + 5t^2 \Rightarrow 5t^2 = 15t$. Since $t 
eq 0$,we get $t = 3 \, s$.Now,substitute $t = 3$ into Equation $2$: $h = 5(3)^2 - 5(3) = 45 - 15 = 30 \, m$.The distance between $P$ and $Q$ is $PQ = 2h - h = h = 30 \, m$. Thus,both $(a)$ and $(b)$ are correct.

The vertical height of point $P$ above the ground is twice that of $Q$. $A$ particle is projected downward with a speed of $5 \, m/s$ from $P$ and at the same time another particle is projected upward with the same speed from $Q$. Both particles reach the ground simultaneously,then:

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