Question

If a ball is thrown vertically upwards with speed $u$, the distance covered during the last $t$ seconds of its ascent is

• A. $ut-\frac{1}{2}g{t}^2$
• B. $\frac{1}{2}g{t}^2$
• C. $ut$
• D. $(u-gt)t$

Answer 1

The correct option is B $\frac{1}{2}g{t}^2$

Let the motion of the ball be as shown in the figure

Here, let the distance covered by the ball in last $t$ seconds be $h$.
So, the velocity of ball at point $B$, $(T-t)\text{sec}$ before (where, $T$ is the time of ascent i.e. $T=\frac{u}{g}$) is given by
$v_B=u-g(T-t)=u-g(\frac{u}{g}-t)=gt$
Thus, we have
$h=v_{B}t-\frac{1}{2}g{t}^2=\left(gt\right)t-\frac{1}{2}g{t}^2=\frac{1}{2}g{t}^2$
Hence, distance covered in last $t$ second is $h=\frac{1}{2}g{t}^2$