Question

A Horizontal rifel is fired at a bull's-eye. The muzzle

Answer 1

Answer:

The horizontal distance is 478.38 m

Solution:

As per the question:

Initial Speed of the bullet in horizontal direction, $v_{x} = 670 m/s$

Initial vertical velocity of the bullet, $v_{y} = 0 m/s$

Vertical distance, y = 0.025 m

Now, for the horizontal distance, 'x':

We first calculate time, t:

$y = v_{y}t - \frac{1}{2}gt^{2}$

(since, motion is vertically downwards under the action of 'g')

$0.025 = 0 - \frac{1}{2}\times 9.8t^{2}$

$t = \sqrt{0.05}{9.8} = 0.0714 s$

Now, the horizontal distance, x:

$x = v_{x}t + \frac{1}{2}a_{x}t^{2}$

$x = v_{x}t + \frac{1}{2}0.t^{2}$

(since, the horizontal acceleration will always be 0)

$x = 670\times 0.714 = 478.38 m$