A car is moving horizontally along a straight line with a uniform velocity of $25{\text{ms}}^{-1}$. A projectile is to be fired from this car in such a way that it will return to it after the car has moved by $100\text{m}$. The speed of the projection relative to car (in $m{s}^{-1}$) must be
(Take $g=10{\text{ms}}^{-2}$)

If the projectile returns back to the car, it must have been projected vertical w.r.t. the car.
$\therefore$ Horizontal velocity of projectile
= horizontal velocity of car
$=25\text{m/s}$
$\therefore$ Time of flight $=\frac{S_x}{V_x}$
$=\frac{100}{25}=4\text{s}$
Now, as projectile reaches car vertical displacement is zero,
Applying $s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$, we get
$0=u_y\left(4\right)+\frac{1}{2}(-10)\left(16\right)$
$80=u_y\left(4\right)$
$\Rightarrow u_y=20\text{m/s}$
Speed of projection relative to the car is $\Rightarrow u_y=20\text{m/s}$