A boy stands at $78.4m$ from a building and throws a ball which just enters a window $39.2m$ above the ground. Calculate the velocity of projection of the ball.
Answer
279k+ views
Hint: 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, ${H_{\max }} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ ……………..$\left( 1 \right)$
Range of projection, $R = \dfrac{{{u^2}\sin 2\theta }}{g}$ ……………..$\left( 2 \right)$
Complete step-by-step solution:
Given:
Range of projection, $R = 78.4m + 78.4m = 156.8m$
Maximum height , ${H_{\max }} = 39.2m$
Take, acceleration due to gravity , $g = 9.8m{s^{ - 2}}$
Using equation $\left( 1 \right)$
That is, ${H_{\max }} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
\[39.2 = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2 \times g}}\] …………… $\left( 3 \right)$
Using equation $\left( 2 \right)$
That is, $R = \dfrac{{{u^2}\sin 2\theta }}{g}$
$156.8 = \dfrac{{{u^2}\sin 2\theta }}{g}$ …………………$\left( 4 \right)$
$156.8 = \dfrac{{{u^2}2\sin \theta \cos \theta }}{g}$ ……………. $\left( 5 \right)$ $\left[ {\because \sin 2\theta = 2\sin \theta \cos \theta } \right]$
Divide equation $\left( 3 \right)$ and equation $\left( 5 \right)$
$\dfrac{{39.2}}{{156.8}} = \dfrac{{\dfrac{{{u^2}{{\sin }^2}\theta }}{{2 \times g}}}}{{\dfrac{{{u^2}2\sin \theta \cos \theta }}{g}}}$
$4 = \dfrac{{\dfrac{{\sin \theta }}{2}}}{{\dfrac{{2\cos \theta }}{1}}}$
Therefore, $\tan \theta = \dfrac{4}{4}$
$\tan \theta = 1$
$\theta = {\tan ^{ - 1}}1$
$\theta = {45^ \circ }$
Substituting in equation $\left( 4 \right)$ we get
$156.8 = \dfrac{{{u^2}\sin 90}}{{9.8}}$
$u = \sqrt {1536.64} $
Therefore, $u = 39.2m{s^{ - 1}}$
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.
Formulas used:
Maximum height, ${H_{\max }} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ ……………..$\left( 1 \right)$
Range of projection, $R = \dfrac{{{u^2}\sin 2\theta }}{g}$ ……………..$\left( 2 \right)$
Complete step-by-step solution:

Given:
Range of projection, $R = 78.4m + 78.4m = 156.8m$
Maximum height , ${H_{\max }} = 39.2m$
Take, acceleration due to gravity , $g = 9.8m{s^{ - 2}}$
Using equation $\left( 1 \right)$
That is, ${H_{\max }} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
\[39.2 = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2 \times g}}\] …………… $\left( 3 \right)$
Using equation $\left( 2 \right)$
That is, $R = \dfrac{{{u^2}\sin 2\theta }}{g}$
$156.8 = \dfrac{{{u^2}\sin 2\theta }}{g}$ …………………$\left( 4 \right)$
$156.8 = \dfrac{{{u^2}2\sin \theta \cos \theta }}{g}$ ……………. $\left( 5 \right)$ $\left[ {\because \sin 2\theta = 2\sin \theta \cos \theta } \right]$
Divide equation $\left( 3 \right)$ and equation $\left( 5 \right)$
$\dfrac{{39.2}}{{156.8}} = \dfrac{{\dfrac{{{u^2}{{\sin }^2}\theta }}{{2 \times g}}}}{{\dfrac{{{u^2}2\sin \theta \cos \theta }}{g}}}$
$4 = \dfrac{{\dfrac{{\sin \theta }}{2}}}{{\dfrac{{2\cos \theta }}{1}}}$
Therefore, $\tan \theta = \dfrac{4}{4}$
$\tan \theta = 1$
$\theta = {\tan ^{ - 1}}1$
$\theta = {45^ \circ }$
Substituting in equation $\left( 4 \right)$ we get
$156.8 = \dfrac{{{u^2}\sin 90}}{{9.8}}$
$u = \sqrt {1536.64} $
Therefore, $u = 39.2m{s^{ - 1}}$
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Define absolute refractive index of a medium

Which of the following would not be a valid reason class 11 biology CBSE

Why should electric field lines never cross each other class 12 physics CBSE

An electrostatic field line is a continuous curve That class 12 physics CBSE

What is meant by monosporic development of female class 11 biology CBSE

Trending doubts
The lightest gas is A nitrogen B helium C oxygen D class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Which place is known as the tea garden of India class 8 social science CBSE

What is pollution? How many types of pollution? Define it

Write a letter to the principal requesting him to grant class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE
