Please clarify my doubt

A shooter aims at a point located on a wall at a distance 5root3 m from him. The gun is at the same level of the point. The bullet would exactly hit the target. But when he fires, with a speed of 10m/s, suddenly gravity disappears. The height above the point where the bullet hits the wall is A) root3 m B) 5/root3 C) 5m D) 5root3 m

Easy Math Editor

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Comments

The shooter accounts for gravity when aiming. Assume that $g = 10 \, m/s^2$ . The shooter first has to determine the appropriate firing angle. The shooting location and the target are at the same vertical position.

Accounting for gravity, the bullet flight time is:

$t_f = \frac{2 v \, sin\theta }{g}$

The horizontal travel distance is:

$x_f = v \, cos \theta \, t_f = \frac{2 v^2 \, sin\theta \, cos \theta }{g}$

Setting this equal to the known horizontal distance to the wall:

$\frac{2 v^2 \, sin\theta \, cos \theta }{g} = d \\ sin\theta \, cos \theta = \frac{g d}{2 v^2} = \frac{10 \times \, 5 \sqrt{3}}{2 \times 10^2} = \frac{\sqrt{3}}{4}$

This yields two possible launch angles:

$\theta = \frac{\pi}{3} \, \text{or} \, \frac{\pi}{6}$

Now we will compute a general formula for the final vertical displacement, assuming gravity disappears right at launch. The total flight time in this case is:

$t_f = \frac{d}{v \, cos\theta}$

The final vertical displacement is:

$y_f = v \, sin\theta \, t_f = \frac{d \, v \, sin\theta}{v \, cos\theta} = d \, tan \theta$

For $\theta = \frac{\pi}{3}$ :

$y_f = d \, tan \Big (\frac{\pi}{3} \Big ) = 5 \, \sqrt{3} \, \sqrt{3} = 15$

For $\theta = \frac{\pi}{6}$ :

$y_f = d \, tan \Big (\frac{\pi}{6} \Big ) = 5 \, \sqrt{3} \, \frac{1}{\sqrt{3}} = 5$

Steven Chase - 2 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$