A ball is thrown vertically upward with a velocity of $10\text{m/s}$ from the top of a tower of height of $50\text{m}$. Find the time taken by the ball to reach a height $10\text{m}$ from the ground. (Take $g=10{\text{m/s}}^2$)

The correct option is C $4\text{s}$
Taking origin $O$ at ground and vertically upward direction as $+y$ axis direction.

Initial velocity of ball:
$u_y=10\text{m/s},u_x=0$
Acceleration of ball:
$a_y=-g=-10{\text{m/s}}^2$, $a_x=0$
With respect to origin $O$, the displacement of ball in $y$-direction when it is at a height $10\text{m}$ from ground is:
$S_y=-40\text{m}$
Using kinematic equation in $y$-direction:
$S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$\Rightarrow -40=10t+\frac{1}{2}(-10)t^2$
$\Rightarrow 5t^2-10t-40=0$
$t^2-2t-8=0$
$\therefore (t-4)(t+2)=0$
$\Rightarrow t=4,-2$
Neglecting -ve value of time, we get:
$t=4\text{s}$