What Angle Should a Sniper Adjust to Hit a Target 1000m Away?

[rogers236]

Homework Statement

A sniper barrel is exactly one meter above the ground and perfectly horizontal. 1000m away is a target one meter above the ground. if the bullet leaves the gun with a muzzle velocity of 1500m/s will it reach the target? (no, next part is my problem) at what angle should the sniper shoot to hit the target?

\Deltax = 1000m
\Deltay = 0m
a_x = 0
a_y = -9.8m/s² (gravity)
V_0x = ?
V_0y = ?
time (t) = ?
angle \theta = ?

V₀ = 1500m/s

Homework Equations

\DeltaX = v*t
\DeltaV = V - V₀ = a*t
\DeltaX = (1/2)*a*t² + V₀*t
V² = 2*a*\DeltaX - V₀²

V_0x = V₀*cos\Theta
V_0y = V₀*sin\Theta

*the first four equations can be used for either the X or Y dimension; \DeltaX will just change to \DeltaY

The Attempt at a Solution

I'm not looking for the answer, just a point in the right direction please.

For part one of the question, I found that it would take the bullet 2/3 of a second to reach the target in the X dimension. But in that same time, it would fall 2.2m, hitting the ground 322.3m from the target.

 

[LowlyPion]

Welcome to PF.

You very much need to consider the components of x,y motion. The time to go up and then down to the target, being the time required to reach the target.

Here are some formulas if you need them:
https://www.physicsforums.com/showpost.php?p=905663&postcount=2

 

[rogers236]

Thanks for the reply. I checked out that link and then fixed my original post a bit, putting in the equations as I learned them.

 

[LowlyPion]

Thanks for the reply. I checked out that link and then fixed my original post a bit, putting in the equations as I learned them.

Now you need to figure the time it will be in the air as a function of the vertical velocity. Up and down.

Then figure the same time as a function of the horizontal velocity and the known distance. These times of course must be equal, so that begins to put you in position to figure out what angle satisfies both conditions.

 

[rogers236]

I've done that and get:

(-2V_0y)/(-9.8) = Time-y
1000/V_0x = Time-x

then to

((-9.8)1000)/V_0x = -2V_0y

((-9.8)1000)/(1500cos\Theta) = -2(1500)sin\Theta

((-9.8)1000) = -2(1500)sin\Theta(1500cos\Theta)

((-9.8)1000) = (-2)(1500)(1500)sin\Thetacos\Theta

((-9.8)(1000))/((-2)(1500)(1500)) = sin\Thetacos\Theta

0.00217777778 = sin\Thetacos\Theta

Now what?

 

[LowlyPion]

You might want to recognize the identity

2SinθCosθ = Sin2θ

It's a terrifically useful trig identity that you may run into again with projectile motion equations.