Solving Two Ballistics Problems

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The discussion revolves around solving two ballistics problems involving projectile motion. The first problem requires calculating how far below a target a bullet will hit when fired horizontally at 225 m/s over a distance of 81 m, with a correction noted for the equation used. The second problem involves a bullet fired at a 15.2° angle from a height of 63.9 m, where the initial vertical velocity must be considered, leading to a recalculated horizontal distance of 201.965 m. Participants emphasize the importance of using the correct equations and accounting for initial velocities in projectile motion. The conversation highlights the need for careful step-by-step verification of calculations to ensure accuracy.
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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 = Vot + 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 = Vot + 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 = Vot + 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 = Vot + 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 = Vot + 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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