The ratio of maximum vertical to maximum horizontal distances travelled by a projectile with \( \theta \) as the angle of initial velocity with ground depends on

Question

A projectile is fired with an initial velocity \( u \) at an angle \( \theta \) to the horizontal. The time of flight is \( T \). What is the maximum height \( H \) reached by the projectile?

Answer 1

Step 1: Write \( H \) and \( R \) directly in terms of the velocity components \( u_x = u\cos\theta \) and \( u_y = u\sin\theta \): \( H = \dfrac{u_y^2}{2g} \), \( R = \dfrac{2u_xu_y}{g} \).
Step 2: Divide: \( \dfrac{H}{R} = \dfrac{u_y^2/2g}{2u_xu_y/g} = \dfrac{u_y}{4u_x} \).
Step 3: Since \( \dfrac{u_y}{u_x} = \tan\theta \), the ratio becomes \( \dfrac{H}{R} = \dfrac{\tan\theta}{4} \), which depends on \( \tan\theta \).
\[ \boxed{\frac{H}{R} \propto \tan\theta} \]

Answer 2

Another way to identify the dependence is to look at extreme angles. As \( \theta \to 90^\circ \) (a nearly vertical throw), the horizontal range \( R \to 0 \) while the maximum height \( H \) stays finite and large, so the ratio \( H/R \to \infty \). Let's see which option's function actually blows up to infinity as \( \theta \to 90^\circ \).

\( \sin\theta \): As \( \theta \to 90^\circ \), \( \sin\theta \to 1 \), a finite value — it does not diverge, so it cannot describe the ratio's behavior near vertical throws.
\( \cos\theta \): As \( \theta \to 90^\circ \), \( \cos\theta \to 0 \), which goes to zero, not infinity — the opposite trend of what's needed.
\( \sin2\theta \): As \( \theta \to 90^\circ \), \( \sin2\theta \to \sin180^\circ = 0 \), again the opposite trend, not diverging.
\( \tan\theta \): As \( \theta \to 90^\circ \), \( \tan\theta \to \infty \), exactly matching the expected divergence of \( H/R \) for a near-vertical throw.

Only \( \tan\theta \) reproduces the correct limiting behavior of the height-to-range ratio as the launch angle approaches 90°.

Therefore, the correct answer is \( \tan\theta \).

Answer 3

Another way to identify the dependence is to look at extreme angles. As \( \theta \to 90^\circ \) (a nearly vertical throw), the horizontal range \( R \to 0 \) while the maximum height \( H \) stays finite and large, so the ratio \( H/R \to \infty \). Let's see which option's function actually blows up to infinity as \( \theta \to 90^\circ \).

\( \sin\theta \): As \( \theta \to 90^\circ \), \( \sin\theta \to 1 \), a finite value — it does not diverge, so it cannot describe the ratio's behavior near vertical throws.
\( \cos\theta \): As \( \theta \to 90^\circ \), \( \cos\theta \to 0 \), which goes to zero, not infinity — the opposite trend of what's needed.
\( \sin2\theta \): As \( \theta \to 90^\circ \), \( \sin2\theta \to \sin180^\circ = 0 \), again the opposite trend, not diverging.
\( \tan\theta \): As \( \theta \to 90^\circ \), \( \tan\theta \to \infty \), exactly matching the expected divergence of \( H/R \) for a near-vertical throw.

Only \( \tan\theta \) reproduces the correct limiting behavior of the height-to-range ratio as the launch angle approaches 90°.

Therefore, the correct answer is \( \tan\theta \).

The ratio of maximum vertical to maximum horizontal distances travelled by a projectile with \( \theta \) as the angle of initial velocity with ground depends on

Show Hint

The Correct Option is D

Approach Solution - 1

Approach Solution -2

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