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

y=-1.66 m

Step-by-step explanation:

We know that

Range R

`R=(u^2sin2\theta )/(g)`

Here given that

u= 275 m/s

R=75 m

Now by putting the values

`R=(u^2sin2\theta )/(g)`

`35=(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-(1)/(2)gt^2`

`y=0.62 * 0.65-(1)/(2)* 9.81* 0.65^2`

y=-1.66 m