Question

A body is projected such that its kinetic energy at the top is (3/4)th of its initial kinetic

energy. What is the angle of projection with the horizontal?

(a) 300

(b) 600

(c)450

(d) 1200​

Answer 1

Solution :

⏭ Given:

✏ A body is projected such that its kinetic energy at the top is 3/4th of itsbinitial kinetic energy.

⏭ To Find:

✏ Angle of projection with horizontal

⏭ Formula derivation:

• At initial point

$\bigstar \sf \: \red{ K_o = \dfrac{1}{2}m {v}^{2} }$

✏ v = initial velocity of projection

• At highest point

$\bigstar \sf \: \blue{K = \dfrac{1}{2}m {v}^{2} { \cos}^{2}\theta} \\ \\ \because \sf \: at \: highest \: point \: \green{v_x = v \cos \theta} \: and \: \purple{v_y = 0}$

_________________________________

$\bigstar \: \underline{ \boxed{ \bold{ \sf{ \pink{ \dfrac{K}{K_o} = { \cos}^{2} \theta}}}}} \: \bigstar$

⏭ Calculation:

$\mapsto \sf \: \dfrac{3}{4} \dfrac{ \cancel{K_o}}{ \cancel{K_o}} = { \cos}^{2} \theta \\ \\ \mapsto \sf \: \cos \theta = \sqrt{ \dfrac{3}{4}} \\ \\ \mapsto \sf \: \cos \theta = \dfrac{ \sqrt{3}}{2} \\ \\ \mapsto \sf \: \theta = { \cos}^{ - 1} \dfrac{ \sqrt{3} }{2} \\ \\ \mapsto \: \underline{ \boxed{ \bold{ \sf{ \gray{ \theta = 30 \degree}}}}} \: \orange{ \bigstar}$

Answer 2

$\Large\boxed{\sf{\quad(a)\quad\!30^o\quad}}$

Solution:-

We consider law of conservation of mechanical energy.

At the point of projection, kinetic energy exists but no potential energy.

• $\sf{K_1=\dfrac{1}{2}mu^2}$

• $\sf{U_1=0}$

And at the top, given that the kinetic energy of the body is $\sf{\left(\dfrac{3}{4}\right)^{th}}$ of its kinetic energy. So,

• $\sf{K_2=\dfrac{3}{4}\,K_1}$

The potential energy of the body at this point is, by energy conservation,

$\longrightarrow\sf{K_1+U_1=K_2+U_2}$

$\longrightarrow\sf{K_1=\dfrac{3}{4}\,K_1+U_2}$

$\longrightarrow\sf{U_2=\dfrac{1}{4}\,K_1}$

That is,

$\longrightarrow\sf{mgh=\dfrac{1}{4}\cdot\dfrac{1}{2}\,mu^2}$

$\longrightarrow\sf{gh=\dfrac{1}{8}\,u^2}$

$\longrightarrow\sf{h=\dfrac{u^2}{8g}\quad\quad\dots(1)}$

But $\sf{h}$ is the maximum height.

$\longrightarrow\sf{h=\dfrac{u^2\sin^2\theta}{2g}}$

Thus (1) becomes,

$\longrightarrow\sf{\dfrac{u^2\sin^2\theta}{2g}=\dfrac{u^2}{8g}}$

$\longrightarrow\sf{\sin^2\theta=\dfrac{1}{4}}$

For acute angle $\theta,$

$\longrightarrow\sf{\sin\theta=\dfrac{1}{2}}$

Therefore,

$\longrightarrow\sf{\underline{\underline{\theta=30^o}}}$