Question

The horizontal range of a projectile is $ 400\,m $ . 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{2 u \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
$H=\frac{u^{2} \sin ^{2} \theta}{2 g}$
$=\frac{u^{2} \sin ^{2} 45^{\circ}}{2 g}$
$=\frac{400}{2} \times \frac{1}{2}=100\, m$