Question

If a gun is sighted to hit targets that are at the same height as the gun and 75 m away at the same height, how low, as a positive number in meters, will the bullet hit if aimed directly at a target 180 m away? The muzzle velocity of the bullet is 275 m/s.

Answer 1

Answer:

y=-1.66 m

Explanation:

   We know that

Range R

$R=\dfrac{u^2sin2\theta }{g}$

Here given that

u= 275 m/s

R=75 m

Now by putting the values

$R=\dfrac{u^2sin2\theta }{g}$

$35=\dfrac{275^2sin2\theta }{9.81}$

θ=0.13°

Now horizontal component of velocity u will be u cosθ.

Horizontal component = 275 cos0.13 = 274.99 m/s

So the time required to cover 180 m in horizontal direction t

180 = 274.99 x t

t=0.65 sec

Now vertical component of velocity u will be u sinθ.

Horizontal component = 275 sin0.13 = 0.62 m/s

So now vertical displacement y will be

$y=u_yt-\dfrac{1}{2}gt^2$

$y=0.62 \times 0.65-\dfrac{1}{2}\times 9.81\times 0.65^2$

y=-1.66 m