- #1
netrunnr
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A shell is shot with initial velocity [tex]\vec{}v_{0}[/tex] is 20ms[tex]^{-1}[/tex] at an angle [tex]\vartheta[/tex] = 60 degrees. At the top of the trajectory, the shell explodes into two fragments of equal mass. Fragment1 has a speed of 0 immediately after the explosion and falls vertically. How far from the gun does fragment2 land assuming terrain is level and there is no air drag?
Equations used:
F = ma
v[tex]^{2}[/tex] = v(of origin)[tex]^{2}[/tex] a + 2a(y-y[tex]_{o}[/tex])
p = mv
Initial middle
p=20ms[tex]-{1}[/tex]m p = [tex]\frac{1}{2}[/tex]mv + [tex]\frac{1}{2}[/tex]m x 0
=> 20m => [tex]\frac{1}{2}[/tex]mv + [tex]\frac{1}{2}[/tex]mx0
=> 20m = [tex]\frac{1}{2}[/tex]mv
=> 20 = [tex]\frac{0.5mv}{m}[/tex] = [tex]\frac{1}{2}[/tex]v = 40ms[tex]-{1}[/tex]
so the middle position is p = 40ms[tex]-{1}[/tex]
breaking this down to x y components:
for x:
cos[tex]\vartheta[/tex] = [tex]\frac{x}{20}[/tex]
cos [tex]\vartheta[/tex] x 20 = x
[tex]\frac{1}{2}[/tex] x 20 = x
x = 10
for y:
sin[tex]\vartheta[/tex] = [tex]\frac{y}{20}[/tex]
sin[tex]\vartheta[/tex] x 20 = y
[tex]\sqrt{\frac{3}{2}}[/tex] x 20 = y
[tex]\frac{20 x [tex]\sqrt{3}[/tex]}{2}[/tex] = 10 x [tex]\sqrt{3}[/tex]
y = 10 x [tex]\sqrt{3}[/tex]
using v[tex]^{2}[/tex] = v[tex]_{o}[/tex][tex]^{2}[/tex] a + 2a(y-y[tex]_{o}[/tex])
0 = (10 x [tex]\sqrt{3}[/tex])[tex]^{2}[/tex] + 2 x 9.8ms[tex]-{1}[/tex](y - 0)
0 = 100 x 3 + 2 x 9.8 x y
y = [tex]\frac{300}{19.6}[/tex] = 15.3
fragment1 dropped from a height of 15.3 meters
I am lost as to what equation to use to calculate the distance fragment 2 traveled from here. I know its simple but somehow I am not able to think of what to do next... I need the distance in the x direction that fragement2 traveled.
hints??
thanx!
Equations used:
F = ma
v[tex]^{2}[/tex] = v(of origin)[tex]^{2}[/tex] a + 2a(y-y[tex]_{o}[/tex])
p = mv
Initial middle
p=20ms[tex]-{1}[/tex]m p = [tex]\frac{1}{2}[/tex]mv + [tex]\frac{1}{2}[/tex]m x 0
=> 20m => [tex]\frac{1}{2}[/tex]mv + [tex]\frac{1}{2}[/tex]mx0
=> 20m = [tex]\frac{1}{2}[/tex]mv
=> 20 = [tex]\frac{0.5mv}{m}[/tex] = [tex]\frac{1}{2}[/tex]v = 40ms[tex]-{1}[/tex]
so the middle position is p = 40ms[tex]-{1}[/tex]
breaking this down to x y components:
for x:
cos[tex]\vartheta[/tex] = [tex]\frac{x}{20}[/tex]
cos [tex]\vartheta[/tex] x 20 = x
[tex]\frac{1}{2}[/tex] x 20 = x
x = 10
for y:
sin[tex]\vartheta[/tex] = [tex]\frac{y}{20}[/tex]
sin[tex]\vartheta[/tex] x 20 = y
[tex]\sqrt{\frac{3}{2}}[/tex] x 20 = y
[tex]\frac{20 x [tex]\sqrt{3}[/tex]}{2}[/tex] = 10 x [tex]\sqrt{3}[/tex]
y = 10 x [tex]\sqrt{3}[/tex]
using v[tex]^{2}[/tex] = v[tex]_{o}[/tex][tex]^{2}[/tex] a + 2a(y-y[tex]_{o}[/tex])
0 = (10 x [tex]\sqrt{3}[/tex])[tex]^{2}[/tex] + 2 x 9.8ms[tex]-{1}[/tex](y - 0)
0 = 100 x 3 + 2 x 9.8 x y
y = [tex]\frac{300}{19.6}[/tex] = 15.3
fragment1 dropped from a height of 15.3 meters
I am lost as to what equation to use to calculate the distance fragment 2 traveled from here. I know its simple but somehow I am not able to think of what to do next... I need the distance in the x direction that fragement2 traveled.
hints??
thanx!