Finding range and height of bullets when given velocity

[AngryApple]

Homework Statement

You are designing a firing range. The range will be 200 yards in length. You must build a backstop that is of sufficient height to contain the projectiles. There are .22 caliber long rifle ammo and .308 caliber center fire ammo used on the range. A shooter discharges a round at a 60 degree angle to the horizontal, but you don't know which of the ammo he used. Calculate how tall the backstop would be to keep the fired round from escaping.

The paperwork says I need to find:

• The range for each type of ammo fired at the angle,
  calculations for the height of each bullet at the time it reaches the end of the range

Homework Equations

Initial velocity of .22 caliber = 330 m/s
Initial velocity of .308 caliber = 860 m/s

The Attempt at a Solution

I'm going to start off working with the .22 caliber bullet, and finding its range.
x = t * V * cosθ
60.96 meters (200 yards) = t * 330m/s * cos(60°)
Solving for t, I get ≈.369 seconds.
Then, I substitute the T value to try and find the range of the bullet.
x = .369 seconds * 330m/s * cos(60)
I get the range of the bullet to be 60.885 meters.

Can someone point me in the right direction to find the range of the bullet? I have a good feeling my way is drastically wrong.

 

[gneill]

Homework Statement

You are designing a firing range. The range will be 200 yards in length. You must build a backstop that is of sufficient height to contain the projectiles. There are .22 caliber long rifle ammo and .308 caliber center fire ammo used on the range. A shooter discharges a round at a 60 degree angle to the horizontal, but you don't know which of the ammo he used. Calculate how tall the backstop would be to keep the fired round from escaping.

The paperwork says I need to find:

• The range for each type of ammo fired at the angle,
  calculations for the height of each bullet at the time it reaches the end of the range

Homework Equations

Initial velocity of .22 caliber = 330 m/s
Initial velocity of .308 caliber = 860 m/s

The Attempt at a Solution

I'm going to start off working with the .22 caliber bullet, and finding its range.
x = t * V * cosθ
60.96 meters (200 yards) = t * 330m/s * cos(60°)
Solving for t, I get ≈.369 seconds.
Then, I substitute the T value to try and find the range of the bullet.
x = .369 seconds * 330m/s * cos(60)
I get the range of the bullet to be 60.885 meters.

Can someone point me in the right direction to find the range of the bullet? I have a good feeling my way is drastically wrong.

Hi AngryApple, Welcome to Physics Forums.

The range of a projectile launched at a given angle is the total horizontal distance it travels (here assuming that it is launched at "ground" level and lands at the same level). So to begin with you'll want to ignore the backstop completely and see how far the bullet will travel unimpeded.

Hint: You'll want to deal with the motion in the vertical direction first in order to see how long it takes to reach the ground.

 

[rude man]

1.Can someone point me in the right direction to find the range of the bullet? I have a good feeling my way is drastically wrong.

You already know the range of the bullet. It's 200 yards!

What you need is the vertical distance of the bullet at that range. Write the equations for x(t) and y(t). You were on the right track to compute T, the time for which x = 200 yds. Now plug that value of T into the y expression to find y(x=200 yds).