Q. A projectile shot into air at some angle with the horizontal has a range of 200 m. If the time of flight is 5 s, then the horizontal component of the velocity of the projectile at the highest point of trajectory is
A) $40\ ms^{- 1}$
B) $0\ ms^{- 1}$
C) $9.8\ ms^{- 1}$
D) Equal to the velocity of projection of the projectile
Explanation
(a) $R = \frac{v^{2}\sin{2\theta}}{g} = 200,\ T = \frac{2v\sin\theta}{g} = 5$
Dividing, $\frac{v^{2} \times 2\sin\theta\cos\theta}{g} \times \frac{g}{2v\sin\theta} = \frac{200}{5} = 40$
or $v\cos\theta = 40ms^{- 1}$
It may be noted here that the horizontal component of the velocity of projection remains the same during the flight of the projectile
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.