Question

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

(Take $g=10m/s^2$)

• A. $5t^2$
• B. $t^2$
• C. $3t^2$
• D. $8t^2$

Answer 1

The correct option is A $5t^2$

Let, the maximum height of flight is $H$ and the time of flight is $T=u/g$

So, $H=uT-\frac{1}{2}g{T}^2$

Now the distance covered by the ball in time $(T-t)$ sec is

$h=u(T-t)-\frac{1}{2}g(T-t{)}^2$

So, the distance covered in last t sec is

$d=H-h=uT-\frac{1}{2}g{T}^2-uT+ut+\frac{1}{2}g{T}^2-gTt+\frac{1}{2}g{t}^2=ut-gTt+\frac{1}{2}g{t}^2=ut-g\left(u/g\right)t+\frac{1}{2}g{t}^2=\frac{1}{2}g{t}^2$