Q. An arrow is projected into air. Its time of flight is 8 s and range 200 m. What is the maximum height reached by it? (Take $g\ = \ 10\ ms^{- 2}$)
A) 31.25 m
B) 24.5 m
C) 18.25 m
D) 46.75 m
Explanation
(a) Given, $5 = \frac{2u\sin\theta}{2g}\ \ or\ \frac{u\sin\theta}{g} = \frac{5}{2}$
Maximum height $= \frac{u^{2}\sin\theta}{2g} = \frac{g}{2}\left( \ \frac{u^{2}\sin^{2}\theta}{g^{2}} \right)$
$= \frac{g}{2} \times \left( \frac{5}{2} \right)^{2} = \frac{10}{2} \times \frac{25}{4} = 31.25\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.