Question

A ball is projected from a point at two different angle with same speed $u$ and land at the same point in the both the cases. Then

• A. The difference between two angles of projection is ${90}^o$
• B. The maximum higher attained by the ball in both cases is equal
• C. The sum of maximum heights for the two cases is $u^2/\left(2g\right)$
• D. The maximum height attained by the ball in one case must be twice of the maximum height attained by the ball in second case

Answer 1

The correct option is C The sum of maximum heights for the two cases is $u^2/\left(2g\right)$

Two projectile have the same range if they are launched at angles which are complementary to each other. That is, if the 1st projectile is launched aat an angle $\theta$ from the horizontal, then the other projectile must be launched at an angle of $\frac{pi}{2}-\theta$ from the horizontal for having the same range

Now maximum height of a projectile $h=\frac{u^2{\sin }^2\theta }{2g}$

For the 1st ball we have $h_1=\frac{u^2{\sin }^2\theta }{2g}$

For the 2nd ball we have $h_2=\frac{u^2{\sin }^2(\frac{\pi }{2}-\theta )}{2g}=\frac{u^2{\cos }^2\theta }{2g}$

$h_1+h_2=\frac{u^2}{2g}({\sin }^2\theta +{\cos }^2\theta )$

$\therefore h_1+h_2=\frac{u^2}{2g}$