Question

(पाठ-4दख At whal distance from the gun the bullet will strike the ground, if a bullet is fired with a initial velocity 250 ms at an angle of 30° with the horizontal​

Answer 1

Answer:

5,412.66 m

Explanation:

First find the Vertical Component for Velocity.

therefore:

V( vertical) = sin(30°) * 250

= 1/2 * 250

= 125 m

Using the formula-

$s = ({v}^{2} - {u}^{2} ) \div 2a$

we know that at the maximum height v=0

so

s=( 0- (125)(125) ) / -20

= 781.25 m

with this we can find the time of flight using

$s = ut + \frac{1}{2} a {t}^{2}$

781.25= 1/2 * 10 * t^2

t= 12.5 s

this is only halfway, so multiply by two

t= 25

therefore the horizontal component is

cos 30° * 250= 125* sqrt 3

therefore distance = s * t

therefore distance= 25 * 125* sqrt 3

= 5412.66 m