# Help please! Tricky-Finding the angle reqd. to hit a target

## Homework Statement

A gun shoots a bullet with a velocity of magnitude 490 m/s. The goal is to hit the target 980 meters away. How high above the target must you aim to correct for gravity? (assume gun and target are same height)

## Homework Equations

I worked the problem by first getting the x- and y- component equations for position and velocity. I got the x-position equation at final time T, 980 meters=490cos(theta)*T, solved for T, plugged it into the y-position equation at T, 0=-1/2(9.8)T^2 + 490sin(theta)T.

## The Attempt at a Solution

Solving for theta, I got about .115 degrees as the answer. The Book says the answer is 1.15 degrees. What did I do wrong? Thank you!

LowlyPion
Homework Helper

## Homework Statement

A gun shoots a bullet with a velocity of magnitude 490 m/s. The goal is to hit the target 980 meters away. How high above the target must you aim to correct for gravity? (assume gun and target are same height)

## Homework Equations

I worked the problem by first getting the x- and y- component equations for position and velocity. I got the x-position equation at final time T, 980 meters=490cos(theta)*T, solved for T, plugged it into the y-position equation at T, 0=-1/2(9.8)T^2 + 490sin(theta)T.

## The Attempt at a Solution

Solving for theta, I got about .115 degrees as the answer. The Book says the answer is 1.15 degrees. What did I do wrong? Thank you!

I think your equations look right. I think they made an error in the final answer. Perhaps they made a simplifying approximation?

Vx = V*Cosθ =490 Cosθ
X = V*t = 980 = 490*Cosθ *t
t = 2/Cosθ

For your Y equation you get that Vy*t = 1/2 g*t2

This yields Vy = V*Sinθ = 1/2 g*t = 1/2 *g *(2/Cosθ ) = g/Cosθ
Solving gives Sinθ Cosθ = Sin(2*θ) = g/490 = 9.8/490 = .02
(Note the identity Sinθ Cosθ = Sin(2*θ))

sin-1(.02) = 1.15 degrees

But that = 2θ,
so I'd say θ = (1.15)/2 degrees = .575 degrees

Vy*t = 1/2 g*t^2 can't be right though, because that leaves out the initial y velocity of the bullet, right? I'm thinking it would have to be Vy*t = 1/2 g*t^2 + V(0)sin(theta)

LowlyPion
Homework Helper
Vy*t = 1/2 g*t^2 can't be right though, because that leaves out the initial y velocity of the bullet, right? I'm thinking it would have to be Vy*t = 1/2 g*t^2 + V(0)sin(theta)

The V*Sinθ term is the Vy term

Oh, ok, I get it now. Thank you very much!