Question

A body is projected with velocity \[u\] such that its horizontal range and maximum vertical heights are the same. The maximum heights is:
A. \[\dfrac{{{u^2}}}{{2g}}\]
B. \[\dfrac{{3{u^2}}}{{4g}}\]
C. \[\dfrac{{16{u^2}}}{{17g}}\]
D. \[\dfrac{{8{u^2}}}{{17g}}\]

Answer 1

Hint: Initially, we write down the given information. Then we equate the formulas for horizontal range and the maximum height to each other. This allows us to find the value of sine of the angle of projection. We apply this value of sine of the angle of projection to the formula of maximum height and find the maximum height of the motion.

Formulas used:
The maximum height of projectile motion is given by the formula,
\[H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}\]
The horizontal range of the projectile motion is given by the formula,
\[R = \dfrac{{{u^2}\sin 2\theta }}{g}\]
Where \[\theta \] is the angle of projection, \[u\] is the initial velocity of the projectile and \[g\] is the acceleration due to gravity.

Complete step by step answer:
It is given that \[u\] is the initial velocity of the projectile and the maximum vertical height the projectile can reach is equal to the horizontal range of the projectile.This gives us,
\[\dfrac{{{u^2}\sin 2\theta }}{g} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}\]
Cancelling the like terms from both sides, we get
\[\sin 2\theta = \dfrac{{{{\sin }^2}\theta }}{2}\]
We simplify this equation and arrive at,
\[2\sin \theta \cos \theta = \dfrac{{\sin \theta \times \sin \theta }}{2}\]
We cancel the sine value and the remainder will be,
\[2\cos \theta = \dfrac{{\sin \theta }}{2}\]
Taking the trigonometric functions to one side and the numerals to the other,
\[4 = \dfrac{{\sin \theta }}{{\cos \theta }}\]

But we know that this is the tangent. That is, \[\tan \theta = 4\].
Using the trigonometric identity, \[1 + {\tan ^2}\theta = {\sec ^2}\theta \].
We can find the value of secant squared and taking the reciprocal, we find the value of cosine squared \[{\sec ^2}\theta = 17\] and we take the reciprocal to get, \[{\cos ^2}\theta = \dfrac{1}{{17}}\]
We apply another trigonometric identity,
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
We get the value of sine squared as,
\[{\sin ^2}\theta = 1 - {\cos ^2}\theta \\
\Rightarrow {\sin ^2}\theta = \dfrac{{17 - 1}}{{17}} \\
\Rightarrow {\sin ^2}\theta = \dfrac{{16}}{{17}}\]
Substituting this value in the equation to find the maximum height, we get
\[H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} \\
\Rightarrow H= \dfrac{{{u^2} \times 16}}{{2 \times 17 \times g}} \\
\therefore H= \dfrac{{8{u^2}}}{{17g}}\]

Hence, the right answer is option D.

Note: In projectile motion, the vertical component of velocity is zero or absent. This is because, as the projectile moves upward the movement taking place will be against gravity and this makes the projectile to decelerate.