Question

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

• A. $2g{t}^2$
• B. $\frac{1}{2}g{t}^2$
• C. $g{t}^2$
• D. $\frac{3}{2}g{t}^2$

Answer 1

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

If ball is thrown with velocity $u$, then time of flight $=\frac{u}{g}$

Velocity after $(\frac{u}{g}-t)sec\Rightarrow v=u-g(\frac{u}{g}-t)=gt$

So, distance in last $tsec$ $\Rightarrow v^2=u^2-2gh\Rightarrow 0^2=(gt{)}^2-2gh\Rightarrow h=\frac{1}{2}g{t}^2$