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?
5%
10%
20%
40%
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: \[ \% \text{increase} = 1.048 - 1 = 0.048 \approx 5\% \]
A particle is projected at an angle of \( 30^\circ \) from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is \( h_0 \) and height traversed in the last second, before it reaches the maximum height, is \( h_1 \). The ratio \( \frac{h_0}{h_1} \) is __________. [Take \( g = 10 \, \text{m/s}^2 \)]