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Question: A ball is projected from \(10m\) away from a wall. If the initial velocity with which the ball can b...

A ball is projected from 10m10m away from a wall. If the initial velocity with which the ball can be projected is 15ms−115m{s^{ - 1}} then, in order to hit the wall at the highest possible point the angle with the horizontal that it must be projected is
(A) 450{45^0}
(B) 300{30^0}
(C) tan⁡−12.25{\tan ^{ - 1}}2.25
(D) tan⁡−11.5{\tan ^{ - 1}}1.5

Explanation

Solution

In order to solve this question, we should know about the motion of the projectile. When a body is projected with some velocity at some angle horizontally, the body covers some maximum height against gravity and covers maximum range later it falls back on ground with the same velocity, we will find the required angle by using basic formulas of projectile motion.
Formula used: If u is the initial velocity of projectile body and θ\theta be the angle with horizontal and H, R be the maximum height and range attained by projectile body then,
H=u2sin⁡2θ2gH = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} Where g is the acceleration due to gravity
R=u2sin⁡2θgR = \dfrac{{{u^2}\sin 2\theta }}{g}

Complete step-by-step solution:
According to the question, we have given that
u=15ms−1u = 15m{s^{ - 1}} Initial velocity of ball
R′=R2=10mR' = \dfrac{R}{2} = 10m Half of the maximum distance the ball will cover to hit wall at highest point
⇒R=20m\Rightarrow R = 20m
g=9.8ms−2g = 9.8m{s^{ - 2}}
Using the formula,
R=u2sin⁡2θgR = \dfrac{{{u^2}\sin 2\theta }}{g} And putting values we get,
⇒20=(15)2sin⁡2θ9.8\Rightarrow 20 = \dfrac{{{{(15)}^2}\sin 2\theta }}{{9.8}}
⇒sin⁡2θ=196225\Rightarrow \sin 2\theta = \dfrac{{196}}{{225}}
⇒2θ=sin⁡−1(0.87)\Rightarrow 2\theta = {\sin ^{ - 1}}(0.87)
And we know, sin⁡−1(0.87)=600{\sin ^{ - 1}}(0.87) = {60^0}
⇒2θ=60\Rightarrow 2\theta = 60
⇒θ=300\Rightarrow \theta = {30^0}
Hence, the correct option is (B) 300{30^0}.

Note: It should be remembered that, at the highest point of vertical hit on the wall then, the horizontal distance up to that point is half the distance of maximum range of projectile motion because after hitting maximum point on wall the ball have to return to ground and have to cover same horizontal distance again and hit the ground with same velocity with which it is projected if there were no wall.