Question

a cricjet ball is thorwn by maximum distance when thrown at an angle of
A) 30°. B) 15°. C) 45°. D) 90°​

Answer 1

The horizontal distance $\sf R$ travelled by a projectile is given by,

$\sf R \: = \: {\large{ \frac{ {u }^{2} \: { \sin}(2 \theta )}{g}}}$

where $\sf u$ is the initial velocity while projection, and $\sf \theta$ is the angle of projection with the horizontal.

Suppose, a/b be a fraction. It's value will be maximum when either $\sf a$ becomes maximum or $\sf b$ becomes minimum.

Here, as in the denominator, $\sf g$ is a constant, there is no minimum value for it. So we have to find the maximum value of $\sf u^2 \: \sin (2 \theta)$

The max value of $\sf \sin \alpha$ is 1, when $\alpha = 90^o$

Hence, for maximum value of $\sf u^2 \:\sin (2 \theta),$

$\sf \sin(2 \theta) \: = 1 \\ \sf \longrightarrow \sin(2 \theta) \: = \sin {90}^{o} \\ \sf \longrightarrow 2 \theta \: = {90}^{o} \\ \sf \longrightarrow \theta \: = {\frac{90}{2}}^{o} \\ \sf \longrightarrow \theta \: = {45}^{o}$

Hence, the answer is,

$\longrightarrow \underline{\underline{\sf{\green{ \theta = 45^o }}}}$