Question

For a given speed of projection, maximum range of
a projectile is R. What will be the maximum height
attained by projectile when it is projected with
double speed by making 60° with vertical?

Answer 1

Explanation:

u is doubled so formula will be

Max. h= 2u^2 sin 2 60*/g

Answer 2

Answer:

Maximum height, $H=\dfrac{3R}{2}$

Explanation:

The maximum range of the projectile is given by :

$R=\dfrac{u^2\ sin2\theta}{g}$

For maximum range of a projectile, $\theta=45$

$R=\dfrac{u^2}{g}$

$u=\sqrt{Rg}$

The maximum height attained by the projectile when it is projected with  double speed is given by :

$H=\dfrac{u^2\ sin^2\theta}{2g}$

$H=\dfrac{(2\sqrt{Rg})^2\ sin^2(60)}{2g}$

$H=\dfrac{3R}{2}$

So, the maximum height attained by the projectile is 3R/2. Hence, this is the required solution.