Question

Two balls with the same mass and initial velocity are projected at different angles in such a way that the maximum height reached by the first ball is 8 times higher than that of the second ball. $ T_1 $ and $ T_2 $ are the total flying times of the first and second ball, respectively, then the ratio of $ T_1 $ and $ T_2 $ is:

• A. \( 2 : 1 \)
• B. \( \sqrt{2} : 1 \)
• C. \( 4 : 1 \)
• D. \( 2\sqrt{2} : 1 \)

Hint:

In projectile motion, the total time of flight and maximum height are related to the angle of projection. For a given initial velocity, the height and time of flight depend on \( \sin^2 \theta \) and \( \sin \theta \), respectively.

Answer 1

The Correct Option is D

To solve the problem, we need to determine the ratio of the total flying times \( T_1 \) and \( T_2 \) of two balls projected at different angles but with the same initial velocity, given that the maximum height of the first ball is 8 times that of the second ball.

1. First, consider the equations for the maximum height (\( H \)) and time of flight (\( T \)) for projectile motion. The maximum height \( H \) can be expressed as:

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

1. where:
   • \(v_i\) is the initial velocity.
   • \(\theta\) is the angle of projection.
   • \(g\) is the acceleration due to gravity.
2. The total time of flight \( T \) is given by:

\(T = \frac{2v_i \sin \theta}{g}\)

1. We know that the height of the first ball, \( H_1 \), is 8 times the height of the second ball, \( H_2 \):

\(\frac{H_1}{H_2} = 8\)

1. substituting the expression for height:

\(\frac{\frac{v_i^2 \sin^2 \theta_1}{2g}}{\frac{v_i^2 \sin^2 \theta_2}{2g}} = 8\)

1. Simplifying, we get:

\(\frac{\sin^2 \theta_1}{\sin^2 \theta_2} = 8\)

1. We need the ratio of the total times of flight:

\(\frac{T_1}{T_2} = \frac{\frac{2v_i \sin \theta_1}{g}}{\frac{2v_i \sin \theta_2}{g}} = \frac{\sin \theta_1}{\sin \theta_2}\)

1. From step 3, we have:

\(\frac{\sin^2 \theta_1}{\sin^2 \theta_2} = 8\)

1. Taking square roots on both sides, we get:

\(\frac{\sin \theta_1}{\sin \theta_2} = \sqrt{8} = 2\sqrt{2}\)

Therefore, the ratio of the total flying times \( T_1 \) and \( T_2 \) is \( 2\sqrt{2} : 1 \).

Answer 2

We know that the maximum height H for projectile motion is given by: \( H = \frac{v^2 \sin^2 \theta}{2g} \),
Where:
- \( v \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

Given that the maximum height for the first ball is 8 times that of the second ball, we can write: \( H_1 = 8H_2 \).
From the equation for maximum height, we can deduce that: \( \frac{v^2 \sin^2 \theta_1}{2g} = 8 \times \frac{v^2 \sin^2 \theta_2}{2g} \).
Simplifying: \( \sin^2 \theta_1 = 8 \sin^2 \theta_2 \).
Thus: \( \sin \theta_1 = \sqrt{8} \sin \theta_2 \).

Now, the total time of flight T for a projectile is given by: \( T = \frac{2v \sin \theta}{g} \).
Using this for both balls: \( T_1 = \frac{2v \sin \theta_1}{g}, \quad T_2 = \frac{2v \sin \theta_2}{g} \).
The ratio of the total flight times \( \frac{T_1}{T_2} \) is: \( \frac{T_1}{T_2} = \frac{2v \sin \theta_1}{2v \sin \theta_2} = \frac{\sin \theta_1}{\sin \theta_2} \).
Substituting \( \sin \theta_1 = \sqrt{8} \sin \theta_2 \): \( \frac{T_1}{T_2} = \sqrt{8} = 2\sqrt{2} \).
Thus, the ratio of \( T_1 \) and \( T_2 \) is \( 2\sqrt{2} : 1 \).

Therefore, the correct answer is Option (4).