Question

Two objects A and B are projected with same velocity at angles \( \theta \) and \( 90-\theta \) respectively with the horizontal. Then the ratio of maximum heights they reached \( \frac{H_A}{H_B} \) is:

• A. \( \tan \theta \)
• B. \( \tan^2 \theta \)
• C. \( 2 \tan \theta \)
• D. \( \cot^2 \theta \)

Hint:

Projectile motion with complementary angles results in the same range but different maximum heights related by the square of the tangent of the angle.

Answer 1

The Correct Option is B

Step 1: Recall the maximum height formula.
For a projectile launched with speed $v$ at angle $\theta$, only the vertical part of the velocity, $v\sin\theta$, decides how high it rises. The peak height is $H = \dfrac{v^2\sin^2\theta}{2g}$.
Step 2: Height of object A.
Object A is thrown at angle $\theta$, so $H_A = \dfrac{v^2\sin^2\theta}{2g}$.
Step 3: Height of object B.
Object B uses the complementary angle $90^\circ - \theta$, so $H_B = \dfrac{v^2\sin^2(90^\circ-\theta)}{2g}$.
Step 4: Simplify the complementary angle.
Since $\sin(90^\circ-\theta) = \cos\theta$, we get $H_B = \dfrac{v^2\cos^2\theta}{2g}$.
Step 5: Take the ratio.
The speeds and $g$ are the same, so they cancel: \[ \frac{H_A}{H_B} = \frac{\sin^2\theta}{\cos^2\theta} \]
Step 6: Write as a single function.
That ratio is simply $\tan^2\theta$.
\[ \boxed{\tan^2\theta} \]