Solving Two Ballistics Problems

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Homework Help Overview

The discussion revolves around two ballistics problems involving projectile motion. The first problem involves a bullet fired horizontally towards a target 81 m away, while the second problem concerns a bullet fired at an angle of 15.2° below the horizontal from a height of 63.9 m.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the use of kinematic equations to solve for the vertical displacement and horizontal distance of the bullets. There are attempts to clarify the correct application of these equations, particularly regarding initial velocities and the inclusion of factors like gravitational acceleration.

Discussion Status

Some participants have provided guidance on correcting the equations used, particularly noting the need for a factor of 1/2 in the kinematic equation. There is ongoing exploration of the calculations, with some participants questioning the accuracy of their results and seeking further verification of their methods.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the amount of direct assistance they can provide to one another. There is also a recognition of potential errors in calculations, prompting requests for detailed breakdowns of the work done.

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Homework Statement



I need help with these 2 problems:

A hunter aims directly at a target (on the same level) 81 m away. If the bullet leaves the gun at a speed of 225 m/s, how far below the target will the bullet hit?

A rifle bullet is fired at an angle of 15.2° below the horizontal with an initial velocity of 154 m/s from the top of a cliff 63.9 m high. How far from the base of the cliff does it strike the level ground below?

Homework Equations





The Attempt at a Solution



Vx = 225 m/s, x = 81 m
Vyo = 0 m/s ?
a = -9.8 m/s/s, t = .36 s (from Vx = x/t)

so I used y = volt + at^2 to find y, but that's not right...

and for the second one

Vx = 148.613 m/s
Vo = -40.377 m/s, a = -9.8 m/s/s, y = -63.9 m
using the same equation, t = 1.221 s
and V = x/t, x= 181.456 m

again, not right. help please.
 
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black_hole said:

Homework Statement



I need help with these 2 problems:

A hunter aims directly at a target (on the same level) 81 m away. If the bullet leaves the gun at a speed of 225 m/s, how far below the target will the bullet hit?

A rifle bullet is fired at an angle of 15.2° below the horizontal with an initial velocity of 154 m/s from the top of a cliff 63.9 m high. How far from the base of the cliff does it strike the level ground below?

Homework Equations




You're just missing a 1/2 in your equation here:
y = volt + at^2

The Attempt at a Solution



Vx = 225 m/s, x = 81 m
Vyo = 0 m/s ?
a = -9.8 m/s/s, t = .36 s (from Vx = x/t)

so I used y = volt + at^2 to find y, but that's not right...

and for the second one

Vx = 148.613 m/s
Vo = -40.377 m/s, a = -9.8 m/s/s, y = -63.9 m
using the same equation, t = 1.221 s
and V = x/t, x= 181.456 m

again, not right. help please.

In the first one, you're just missing a 1/2 in your equation here:
y = volt + at^2
 
berkeman said:
In the first one, you're just missing a 1/2 in your equation here:

Wow, how embarrassing. However, in the 2nd one. I now get 201.965 m, which is not right?
 
black_hole said:
Wow, how embarrassing. However, in the 2nd one. I now get 201.965 m, which is not right?

Could you please show your new work for the 2nd one?
 
well, using the correct equation y = Vo + 1/2at^2 and keeping everything the same
Vx = 148.613 m/s
Vo = -40.377 m/s, a = -9.8 m/s/s, y = -63.9 m
I found t = 1.359 s
and V = x/t, x= 201.965 m...
 
black_hole said:
y = Vo + 1/2at^2

The bullet in the 2nd problem is fired downward, so it's initial y velocity is not zero...
 
black_hole said:
well, using the correct equation y = Vo + 1/2at^2 and keeping everything the same
Vx = 148.613 m/s
Vo = -40.377 m/s, a = -9.8 m/s/s, y = -63.9 m
I found t = 1.359 s
and V = x/t, x= 201.965 m...

I meant y = volt + 1/2at^2
I'm sorry if I was not clear, but Vyo = -40.377 m/s
 
That looks mostly right then, unless there is a small rounding error or something. Is it not the right answer? If not, could you please write out each step again, and I;ll do a more careful check of the math. (I'm bailing for a couple hours though -- will try to check back a bit later)
 

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