A marksman is trying to hit a target from 1610 meters away. The rifle barrel and target are the same height above the ground the muzzle velocity of the rifle is 1250m/s

Question

1. At what angle should the shooter aim her rifle in order to hit the target?
2. What is the maximum height of the bullet above the rifle barrel?
3. How long does it take the bullet to reach the target?

Answer 1

1.Remember that the horizontal and veritical share time.

2. Handy trig to know: Sine (theta) divided by cosine theta = tan theta. Also sin(theta)cos(theta)= sin(2theta)/2

3. Look at the first part, when it launches to peak height because velocity in vertical is zero at peak. And at peak, the horizintal distance is half, Xhalf =1610/2= 805m

With all that said, now look at the problem. When you have two equations, two unknowns it becomes a substation problem.

a). Horizontal: Vx=Vcos(theta) and Vx= 805m/time. Vertical: Vfy= Viy+gt and Viy=Vsin(theta)

Rearrange Vx for time and add in cosine... t=805m/Vcos(theta).

Look at the vertical: Vfy=0. So the equation becomes 0=Vsine(theta)+gt

Now you can substitute time equation in horizontal into vertical.

0=Vsin(theta)+g(X/VCos(theta)).

Vsin(theta)= gx/(Vcos(theta))

Sin(theta)cos(theta) = gx/V²

remember #2 above

Sin(2theta)/2= gx/V²

now you should be able solve for theta, the angle.

Once you have this, it should be answer to solve b &c.

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