Question

at what angle with the horizontal should a ball be thrown so that its range is related to
the time of flight as R=5T^2

Answer 1

it is one of the simple question just find the angles and find the answer

Answer 2

Explanation:

Given that,

The relationship between the time of flight and horizontal range is given by:

$R=5T^2$.....(1)

The range of projectile is given by :

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

The time of flight is given by :

$T=\dfrac{2v\ sin\theta}{g}$

$\dfrac{v^2\ sin2\theta}{g}=5\times (\dfrac{2v\ sin\theta}{g})^2$

$sin2\theta=\dfrac{20\ sin^2\theta}{g}$

On solving above equations :

$tan\theta=\dfrac{g}{10}$

Since, $g=10\ m/s^2$

$\theta=45^{\circ}$

So, the angle with which the ball is thrown is 45 degree celcius. Hence, this is the required solution.