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Question: A boy stands at \(78.4m\) from a building and throws a ball which just enters a window \(39.2m\) abo...

A boy stands at 78.4m78.4m from a building and throws a ball which just enters a window 39.2m39.2m above the ground. Calculate the velocity of projection of the ball.

Explanation

Solution

In the question they have given maximum height and range of projection for the body which is there in projectile motion. By using the given data we will find the angle of projection then substituting in the equation of range of projection we will find the initial velocity of the body.

Formulas used:
Maximum height, Hmax=u2sin2θ2g{H_{\max }} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} ……………..(1)\left( 1 \right)
Range of projection, R=u2sin2θgR = \dfrac{{{u^2}\sin 2\theta }}{g} ……………..(2)\left( 2 \right)

Complete step-by-step solution:

Given:
Range of projection, R=78.4m+78.4m=156.8mR = 78.4m + 78.4m = 156.8m
Maximum height , Hmax=39.2m{H_{\max }} = 39.2m
Take, acceleration due to gravity , g=9.8ms2g = 9.8m{s^{ - 2}}
Using equation (1)\left( 1 \right)
That is, Hmax=u2sin2θ2g{H_{\max }} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}
39.2=u2sin2θ2×g39.2 = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2 \times g}} …………… (3)\left( 3 \right)
Using equation (2)\left( 2 \right)
That is, R=u2sin2θgR = \dfrac{{{u^2}\sin 2\theta }}{g}
156.8=u2sin2θg156.8 = \dfrac{{{u^2}\sin 2\theta }}{g} …………………(4)\left( 4 \right)
156.8=u22sinθcosθg156.8 = \dfrac{{{u^2}2\sin \theta \cos \theta }}{g} ……………. (5)\left( 5 \right) [sin2θ=2sinθcosθ]\left[ {\because \sin 2\theta = 2\sin \theta \cos \theta } \right]
Divide equation (3)\left( 3 \right) and equation (5)\left( 5 \right)
39.2156.8=u2sin2θ2×gu22sinθcosθg\dfrac{{39.2}}{{156.8}} = \dfrac{{\dfrac{{{u^2}{{\sin }^2}\theta }}{{2 \times g}}}}{{\dfrac{{{u^2}2\sin \theta \cos \theta }}{g}}}
4=sinθ22cosθ14 = \dfrac{{\dfrac{{\sin \theta }}{2}}}{{\dfrac{{2\cos \theta }}{1}}}
Therefore, tanθ=44\tan \theta = \dfrac{4}{4}
tanθ=1\tan \theta = 1
θ=tan11\theta = {\tan ^{ - 1}}1
θ=45\theta = {45^ \circ }
Substituting in equation (4)\left( 4 \right) we get
156.8=u2sin909.8156.8 = \dfrac{{{u^2}\sin 90}}{{9.8}}
u=1536.64u = \sqrt {1536.64}
Therefore, u=39.2ms1u = 39.2m{s^{ - 1}}

Note: Projectile motion is the form of motion experienced by a launched body that is motion of a body which is projected or thrown into the air with the angle made by the object with respect to the ground or x-axis.