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Question: In oblique projectile motion, for maximum horizontal range, the projectile must be thrown at angle o...

In oblique projectile motion, for maximum horizontal range, the projectile must be thrown at angle of
A. 3030^\circ
B. 6060^\circ
C. 4545^\circ
D. 7070^\circ

Explanation

Solution

Recall the formula for horizontal range of projectile in oblique projectile motion. Check from this equation that the horizontal range of the projectile depends on which factors. Hence, determine the angle of projection for which the horizontal range of projectile in oblique projectile motion will be maximum.

Formula used:
The horizontal range RR of a projectile in oblique projectile motion is
R=u2sin2θgR = \dfrac{{{u^2}\sin 2\theta }}{g} …… (1)
Here, uu is the velocity of projection of the projectile, θ\theta is the angle of projection and gg is acceleration due to gravity.

Complete step by step answer:
We have given that the projectile is performing oblique projectile motion. We have asked to determine the angle of projection at which the horizontal range of the projectile is maximum. From equation (1) for the horizontal range of the projectile, it can be concluded that the horizontal range of projectile depends on velocity of projection of the projectile, angle of projectile and acceleration due to gravity. The velocity of projection for projectile and acceleration due to gravity are constant.

For the horizontal range of the projectile to be maximum, the value of sine of the angle in the formula must be equal to one.
sin2θ=1\sin 2\theta = 1
Let us determine the value of angle of projection for maximum horizontal range in the oblique projectile motion. Take sine inverse on both sides of the above equation.
sin1(sin2θ)=sin1(1){\sin ^{ - 1}}\left( {sin2\theta } \right) = {\sin ^{ - 1}}\left( 1 \right)
2θ=sin1(1)\Rightarrow 2\theta = {\sin ^{ - 1}}\left( 1 \right)
θ=902\Rightarrow \theta = \dfrac{{90^\circ }}{2}
θ=45\therefore \theta = 45^\circ
Therefore, the angle of projection for which the horizontal range will be maximum is 4545^\circ .

Hence, the correct option is C.

Note: The students should use the correct formula for horizontal range of the projectile in oblique projectile motion. If this formula horizontal range of the projectile is not taken correctly, the final value of angle of projection will be other than the value we determined and hence will be incorrect.