Question

The range of a projectile that is launched with an initial velocity v at an angle of a with the horizontal is given by R

Answer 1

Answer:

$\theta=30.285^{\circ}$

Step-by-step explanation:

The range of a projectile is given by :

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

Put R = 20 m, u = 15 m/s and finding the value of angle of projection

So,

$R=\dfrac{u^2\sin2\theta}{g}\\\\\sin2\theta=\dfrac{Rg}{u^2}\\\\\sin2\theta=\dfrac{20\times 9.8}{15^2}\\\\\sin2\theta=0.871\\\\2\theta=\sin^{-1}(0.871)\\\\2\theta=60.57\\\\\theta=30.285^{\circ}$

So, the required angle of projection is equal to $30.285^{\circ}$.