Q. If time of flight of a projectile is 10 seconds. Range is 500 $meters$. The maximum height attained by it will be
A) $125\ m$
B) $50\ m$
C) $100\ m$
D) $150\ m$
Explanation
(a) $T = \frac{2u\sin\theta}{g} = 10\ Sec\ \Rightarrow u\sin\theta = 50\ m/s$
$\therefore H = \frac{u^{2}\sin^{2}\theta}{2g} = \frac{\left( u\sin\theta \right)^{2}}{2g} = \frac{50 \times 50}{2 \times 10} = 125\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.