Question

For a projectile the ratio of maximum height reached to the square of flight time is: $(g=10m{s}^{-2})$

• A. 5 : 4
• B. 5 : 2
• C. 5 : 1
• D. 10 : 1

Answer 1

Answer: (A)

Height of projectile $H_{\max }=\frac{u^2{\sin }^2\theta }{2g}$ Time of flight $T=\frac{2u\sin \theta }{g}$ $\therefore$ $\frac{H_{\max }}{T^2}=\frac{u^2{\sin }^2\theta /2g}{{(2u\sin \theta /g)}^2}$ $=\frac{g}{g}=\frac{10}{8}$ $\therefore$ $H_{\max }:T^2=5:4$