Question

What are the two angles of projection of a projectile projected with velocity 30m/s, so that the horizontal range is 45m. Take, g= 10m/s?.

Answer 1

So, the two Angles R :- 15° and 75° .

Answer 2

Answer:

$\theta=15^o, 75^o$

Explanation:

The initial velocity of the projectile is u=30 m/s

The horizontal range is R=45 m

We know that the formulation of range is

$R=\dfrac{u^2 \sin 2\theta}{g}\\\sin 2 \theta=\dfrac{Rg}{u^2}\\\sin 2 \theta=\dfrac{45\times 10}{30^2}\\\sin 2 \theta=0.5\\2 \theta=\sin^{-1} 0.5\\2 \theta=30\\\theta=\dfrac{30}{2}\\\theta=15^o$

The second angle is

$90-\theta\\90-15\\75^o$