Question

Ratio of maximum height to range is?

Answer 1

GiveN :-

• The ratio of kinetic energy at the point of projection to the point of maximum height is 4 : 1 .

To FinD :-

• The ratio of maximum height to its range .

SolutioN :-

Here the ratio of kinetic energy at the point of projection to the point of maximum height is 4 : 1 . We need to find the ratio of maximum height to its range .

We know that in the projectile motion at the highest point the component of velocity is $\bf v \cos\theta$ . And at the initial point is v .

$\red{\frak{ Kinetic\ Energy }}\begin{cases} \sf At \ maximum \ height \ = \ \textsf{\textbf{v cos}}\theta . \\\textsf {At point of projection = \textbf{v}.} \end{cases}$

$\purple{\bigstar}\underline{\boldsymbol{ According\ to \ Question :- }}$

$\sf:\implies\pink{ \dfrac{ Kinetic \ Energy_{(At\ initial \ point )} }{Kinetic\ Energy _{(Maximum\ height )}}=\dfrac{ 4}{1}}\\\\\sf:\implies \dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}m (v cos\theta )^2}= \dfrac{4}{1}\\\\\sf:\implies \dfrac{v^2}{v^2 cos^2\theta }=\dfrac{4}{1}\\\\\sf:\implies \dfrac{1}{cos^2\theta }=\dfrac{4}{1}\\\\\sf:\implies cos^2\theta = \dfrac{1}{4} \\\\\sf:\implies cos\theta =\sqrt{\dfrac{1}{4}}\\\\\sf:\implies cos\theta =\dfrac{1}{2}\\\\\sf:\implies cos \theta = cos 60^o \\\\\sf:\implies\boxed{\pink{\frak{ \theta = 60^{\circ}}}}$

$\rule{200}2$

$\dag \underline{\textsf{\purple{ Finding the ratio of \textbf{Range } and \textbf{ Maximum Height}.}}}$

$\sf:\implies \pink{\dfrac{Maximum\ Height}{Range}=\dfrac{\frac{u^2 sin^2 \theta }{2g}}{\frac{2u ^2sin\theta cos\theta}{g}}}\\\\\sf:\implies \dfrac{H_{(max)}}{R}=\dfrac{sin\theta}{4 cos\theta}\\\\\sf:\implies \dfrac{H_{(max)}}{R}=\dfrac{tan\theta}{4}\\\\\sf:\implies \dfrac{H_{(max)}}{R}= \dfrac{tan 60^o}{4}\\\\\sf:\implies \underset{\blue{\sf Required\ Ratio }}{\underbrace{ \boxed{\pink{\frak{ \dfrac{Height_{(max)}}{Range}= \dfrac{\sqrt3}{4}}}}}}$