Solveeit Logo
Login

Question

Question: The ceiling of a hall is \[40m\] high. For maximum horizontal distance, the angle at which the ball ...

The ceiling of a hall is 40m40m high. For maximum horizontal distance, the angle at which the ball can be thrown with the speed of 56ms−156m{s^{ - 1}} without hitting the ceiling of the hall is (take g=9.8m/s2g = 9.8m/{s^2} )
(A) 25OC{25^{\rm O}}C
(B) 30OC{30^{\rm O}}C
(C) 45OC{45^{\rm O}}C
(D) 60OC{60^{\rm O}}C

Explanation

Solution

We know that using the third equation of motion V2−U2=2as{V^2} - {U^2} = 2as we can write the given relation as V2sin⁡2θ−2gH=0{V^2}{\sin ^2}\theta - 2gH = 0. We will put the value of V, g, H in the equation and find the value of the angle for the maximum horizontal distance.

Complete step by step answer:
It is given in the question that the ceiling of a hall is 40m40m high. Then we have to find the angle at which the ball can be thrown with the speed 56ms−156m{s^{ - 1}} so that it will cover the maximum horizontal distance.
We know that according to the third equation of motion V2−U2=2as{V^2} - {U^2} = 2as. Here the initial velocity of the ball is considered as 0ms−10m{s^{ - 1}} and the distance S is given by H and acceleration is given by g. then we can rewrite this equation as V2sin⁡2θ−2gH=0{V^2}{\sin ^2}\theta - 2gH = 0.
56×56sin⁡2θ−2×9.8×40=056 \times 56{\sin ^2}\theta - 2 \times 9.8 \times 40 = 0
56×56sin⁡2θ=2×9.8×4056 \times 56{\sin ^2}\theta = 2 \times 9.8 \times 40
sin⁡2θ=2×9.8×4056×56{\sin ^2}\theta = \dfrac{{2 \times 9.8 \times 40}}{{56 \times 56}}
sin⁡2θ=7843136{\sin ^2}\theta = \dfrac{{784}}{{3136}}
sin⁡2θ=0.25{\sin ^2}\theta = 0.25
On finding the square root of both the sides we get-
sin⁡θ=0.50\sin \theta = 0.50
We know that the value of sin⁡θ=0.50\sin \theta = 0.50 an angle 30O{30^{\rm O}}.
Therefore, the maximum value θ\theta will be 30O{30^{\rm O}}.

Thus, option B is correct.

Additional information:
It is Note:d that in an open field we want to cover the maximum distance then we have to project our ball at an angle of 45O{45^{\rm O}}. We will get the maximum range 45O{45^{\rm O}}.

Note:
It is Note:d that the majority one does mistake in writing the relation of maximum height, they may merge the relation of range and height, to avoid such mistake we can use these equations of motion to derive the exact relation and solve our problem accordingly.