Question

For an object thrown at 45° to the horizontal, the maximum height H and horizontal range R are related as

Moderate

Solution

As we know that, maximum height of a projectile is given by
$H_{\max} = \frac{u^{2} \sin^{2} θ}{2 g}$
where, u = initial velocity of projectile
g = acceleration due to gravity
and $θ$ = angle of projection.
As per question,
$H_{\max} = \frac{v^{2} \sin^{2} 45^{0}}{2 g}$ ... (i) (As, u = v, $θ$ = 45°)
Now, range of a projectile is given by
$R = \frac{u^{2} \sin 2 θ}{g}$
$\Rightarrow R = \frac{v^{2} \sin (2 \times 45^{0})}{g}$
$\Rightarrow R = \frac{v^{2} \sin 90^{0}}{g}$ ... (ii)
On dividing Eq. (i) by Eq. (ii), we get
$\frac{H_{\max}}{R} = \frac{v^{2} \sin^{2} 45 ° \times g}{2 g \times v^{2} \sin 90 °} = \frac{1}{4 \times 1}$
$\Rightarrow R = 4 H_{\max} = 4 H$

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