Question

Two objects are projected at an angle θ° and (90-θ)°, to the horizontal with the same speed. The ratio of their maximum vertical height is

• A. 1 : Tan θ
• B. 1 : 1
• C. tan² θ : 1
• D. tan θ : 1

Answer 1

The Correct Option is C

For the object projected at an angle \( \theta \):

• Initial vertical velocity \( u_{y1} = u \sin \theta \) 
• Initial horizontal velocity \( u_{x1} = u \cos \theta \)

For the object projected at an angle \( (90 - \theta) \):

• Initial vertical velocity \( u_{y2} = u \sin(90 - \theta) = u \cos \theta \)
• Initial horizontal velocity \( u_{x2} = u \sin(90 - \theta) = u \sin \theta \)

We can calculate the maximum vertical height reached by each object using the vertical displacement equation:

\( h_{\text{max}} = \frac{u_y^2}{2g} \)

For the object projected at angle \( \theta \):

\( h_{\text{max1}} = \frac{(u \sin \theta)^2}{2g} = \frac{u^2 \sin^2 \theta}{2g} \)

For the object projected at angle \( (90 - \theta) \):

\( h_{\text{max2}} = \frac{(u \cos \theta)^2}{2g} = \frac{u^2 \cos^2 \theta}{2g} \)

Now, to find the ratio of their maximum vertical heights, we divide \( h_{\text{max1}} \) by \( h_{\text{max2}} \):

\[ \frac{h_{\text{max1}}}{h_{\text{max2}}} = \frac{\frac{u^2 \sin^2 \theta}{2g}}{\frac{u^2 \cos^2 \theta}{2g}} = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta \]

Therefore, the ratio of their maximum vertical heights is: \( \tan^2 \theta : 1 \) (option C).

Answer 2

To determine the ratio of the maximum vertical heights reached by the two objects, we can analyze their projectile motion.

Assume the initial speed of both objects is \( v \), and they are projected at angles \( \theta^\circ \) and \( (90^\circ - \theta)^\circ \) to the horizontal. 

The maximum vertical height reached by a projectile can be calculated using the formula:

\( H = \frac{v^2 \sin^2(\theta)}{2g} \)

Where:

• H is the maximum vertical height
• v is the initial speed
• \( \theta \) is the launch angle
• g is the acceleration due to gravity

For the first object projected at an angle \( \theta^\circ \):

\( H_1 = \frac{v^2 \sin^2(\theta)}{2g} \)

For the second object projected at an angle \( (90^\circ - \theta)^\circ \):

\( H_2 = \frac{v^2 \sin^2(90^\circ - \theta)}{2g} = \frac{v^2 \cos^2(\theta)}{2g} \)

To find the ratio of the maximum vertical heights, we divide \( H_1 \) by \( H_2 \):

\[ \frac{H_1}{H_2} = \frac{\frac{v^2 \sin^2(\theta)}{2g}}{\frac{v^2 \cos^2(\theta)}{2g}} = \frac{\sin^2(\theta)}{\cos^2(\theta)} = \tan^2(\theta) \]

Therefore, the ratio of the maximum vertical heights reached by the two objects is: \( \tan^2(\theta) : 1 \).

Thus, the correct answer is: \( \tan^2(\theta) : 1 \).