Question

The maximum height attained by the projectile is increased by 10% by keeping the angle of projection constant. What is the percentage increase in the time of flight?

• A. 5%
• B. 10%
• C. 20%
• D. 40%

Hint:

For projectile motion, the time of flight depends on the velocity and angle. The increase in time is less significant than the increase in height because time is linearly proportional to velocity, while height is proportional to the square of velocity.

Answer 1

The Correct Option is A

Step 1: Analyze the relationship between height and time of flight For projectile motion, the maximum height \( H \) and time of flight \( T \) are related to the initial velocity \( u \) and the angle of projection \( \theta \) by the formulas: \[ H = \frac{u^2 \sin^2 \theta}{2g} \] \[ T = \frac{2u \sin \theta}{g} \] Now, if the maximum height increases by 10%, we have: \[ H_2 = 1.1 H_1 \] Since the height is proportional to \( u^2 \), and the time of flight is proportional to \( u \), the time of flight will change as: \[ T_2 = \sqrt{1.1} T_1 \approx 1.048 T_1 \] Thus, the percentage increase in the time of flight is: \[ \% \text{increase} = 1.048 - 1 = 0.048 \approx 5\% \]

Answer 2

Given:
The maximum height of a projectile is increased by 10% by keeping the angle of projection constant.

Step 1: Recall the formula for maximum height \( H \):
\[ H = \frac{u^2 \sin^2 \theta}{2g} \] where \( u \) is the initial velocity, and \( \theta \) is the angle of projection.

Step 2: If \( H \) is increased by 10%, then:
\[ H_{\text{new}} = 1.1 H = \frac{u_{\text{new}}^2 \sin^2 \theta}{2g} \] \[ \Rightarrow u_{\text{new}}^2 = 1.1 u^2 \implies u_{\text{new}} = u \sqrt{1.1} \]

Step 3: Time of flight \( T \) is given by:
\[ T = \frac{2u \sin \theta}{g} \] Similarly, the new time of flight:
\[ T_{\text{new}} = \frac{2 u_{\text{new}} \sin \theta}{g} = \frac{2 u \sqrt{1.1} \sin \theta}{g} = T \sqrt{1.1} \]

Step 4: Calculate the percentage increase in time of flight:
\[ \% \text{ increase} = \left(\sqrt{1.1} - 1\right) \times 100 \approx (1.0488 - 1) \times 100 = 4.88\% \approx 5\% \]

Therefore, the percentage increase in time of flight is:
\[ \boxed{5\%} \]