Projectile motion (1 Viewer)

[weirdguy99]

Need help with this question.

A sniper is standing with his gun raised 30 degrees above the horizontal with the end of the barrel 2.0 metres above the ground. The muzzle velocity of the rifle is 800 m/s. Calculate the range of the bullet.

I think I got the answer but im not really sure since I dont have the answers.

I could easily do the question if it wasnt 2.0 metres above the ground, but since it is, im not sure how to find the extra range reached by the bullet, even if it's very small.

 

[xV1P3R]

Let 2 m above ground level to be "zero" and up to be positive. So the ground has a displacement of -2m

Muzzle velocity : 800m/s
At 30 degrees , vertical velocity = 800sin30 = 400m/s

You want time of flight, so the displacement you want is -2
Looking at just the vertical component:
s(y) = -2 a = -9.8m/s² u(y) = 400m/s t = ?
s(y) = u(y)t + 0.5a(y)t²
-2 = 400t + -4.9t²

Essentially you solve the quadratic, you disregard the negative solution.
You work out the horizontal velocity = 800cos30 = 400sqrt(3)

With your t value that you found before, you just plug it into this equation v(x) = u(x)t giving you the range

 

[xXmuffin0manXx]

Using the quadatic formula
t = 81.63765276
Using delta x = Uxt
= (692.820323)(81.63765276)
= 56560.22496m
= 56600m (3 sig figs)

lol..that's gotta be wrong

 

[shady145]

Using the quadatic formula
t = 81.63765276
Using delta x = Uxt
= (692.820323)(81.63765276)
= 56560.22496m
= 56600m (3 sig figs)

lol..that's gotta be wrong

i got
56560.22495m LOL... yea our answer are correct... we used the right method... so its right enough

 

[fatmike93]

i got the same as you muffin man Don't know what viper's doin, although he could be right ?!

 

[weirdguy99]

Awesome, I got it right Thanks for your input guys.