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Skeet shooting

  • Thread starter joemama69
  • Start date
399
0
1. Homework Statement

A 0.25 - kg skeet is fired at an angle of 300 to the horizontal with a speed of 30 m/s. When it reaches the maximum height, it is hit from below by a 15-g pellet traveling vertically upward at a speed of 200 m/s. The pellet is embedded in the skeet.


a) How much higher does the skeet go up?

b) How much extra distance x, does the skeet travel because of the collision?



2. Homework Equations



3. The Attempt at a Solution

ms = .25. vs = 30... mp = 15000. vp = 200..Q=30 degrees

msvs + mpvp = (ms + mp)v...15000(200) = (.25 + 15000)v
v = 200 = vy'

vx = 30cos30 = vx'=25.98

t = x/30cos30 = x/25.98

y = yo+vy't - .5gt2 = yo + 200(x/25.98) - .5g(x/25.98)2 = yo + 8.25x - x2/.01 ...x = .077 is this correct so far
 

Delphi51

Homework Helper
3,407
10
I don't follow your work for the vertical velocity after the collision.
momentum before = momentum after collision in vertical direction
mv = mv
.015*200 = (.015+.25)v
solve for v.
 
399
0
the pellet is 15g = 15000 kg
 
179
0
Ask yourself if that makes sense. A bullet hitting a skeet will not make the skeet travel at the same velocity as the bullet. It will increase its velocity but not by that much. Looking at the mass of the pellet, does it make sense that it's mass is 15000kg? No, a gram is 1/1000 of a kilogram, not 1000kgs.
 
399
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sorry i always do that

ms = .25. vs = 30... mp = .015. vp = 200..Q=30 degrees

msvs + mpvp = (ms + mp)v... .015(200) = (.25 + .015)v
v = 11.32 = vy'

vx = 30cos30 = vx'=25.98

t = x/30cos30 = x/25.98

y = yo+vy't - .5gt2 = yo + 11.32(x/25.98) - .5g(x/25.98)2 = yo + .44x - (x^2)/.01 ...
 

Delphi51

Homework Helper
3,407
10
v = 11.3 is correct. I don't understand why your last line y= formula has x's in it. The horizontal motion does not affect the vertical motion.
 
399
0
x = vxt.....t = x/vx

isnt that how u typical solve these problems....
 
179
0
You don't need to do that for this problem. You can use the quadratic formula to find t in your y = equation. Since the question asks for how much higher and farther the skeet travels I would suggest first finding the height and distance when it isn't hit, then finding height and distance when it is hit.
 
399
0
ok so y = yo + vyt - .5gt2 = yo + 11.32t - .5(9.8)t2....ymax @ t = 1.16

so y = yo+6.54

x = xo + 25.98(2.31) = xo + 60.01

is this correct, do i need to find the initials before i can solve the quadratic or can i do as i have done and say it is zero and then add it back in after i find t
 

Delphi51

Homework Helper
3,407
10
The 6.54 looks good for part (a).
The 1.16 seconds must be the time of maximum height. I don't know if that will help you find (b). For (b) you are starting over with a new trajectory problem with Vy = 11.3. You'll have to find the x and y coordinates at that point where the pellet hits it and the Vx left from the initial shot. Then you can do the horizontal and vertical formulas to find the horizontal distance.
 

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