Question

The horizontal range of a projectile is $400m$ . The maximum height attained by it will be:

• A. 100 m
• B. 200 m
• C. 400 m
• D. 800 m

Answer 1

Answer: (A)

The horizontal range = horizontal velocity $\times$ time

$R=u_x\times T$
$R=(u\cos \theta )\times \frac{2u\sin \theta }{g}$
$R=\frac{u^{2}2\sin \theta \cos \theta }{g}$
$R=\frac{u^2\sin 2\theta }{g}$
For maximum horizontal range $\sin 2\theta =1$
$\therefore \theta =45^{\circ }$
$\therefore 400=\frac{u^2}{g}$
Also, maximum height of projectile
$H=\frac{u^2{\sin }^2\theta }{2g}$
$=\frac{u^2{\sin }^{2}45^{\circ }}{2g}$
$=\frac{400}{2}\times \frac{1}{2}=100m$