Question

A ball is projected obliquely with a velocity $49m{s}^{-1}$ strikes the ground at a distance of $245m$ from the point of projection. It remained in air for:

• A. $10sec$
• B. $5\sqrt{2}sec$
• C. $3sec$
• D. $2.5sec$

Answer 1

The correct option is B $5\sqrt{2}sec$

The ball is projected obliquely so suppose of projection from horizontal was $\theta$

Then the horizontal range should be $R=\frac{u^{2}Sin2\theta }{g}$

As given in the question $R=245m$ and $u=49m/s$

putting above values in the expression of $R$ we get $Sin2\theta =1$ so $2\theta ={90}^0$ or $\theta ={45}^0$ so $Sin\theta =\frac{1}{\sqrt[2]{22}}$

So the time in air or the time of flight will be $T=\frac{2uSin\theta }{g}=\frac{2\times 49Sin\theta }{9.8}=5\sqrt[2]{22}Sec$

Option B is correct.