A 5.0g bullet is fired from a gun straight up with a velocity of 300. m.s. Neglecting air resistance, how high will the bullet rise?
Pliz show the steps
Pliz show the steps
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kinetic energy = potential energy
1/2 mv^2 = mgh
m=0.005 kg v=300 m/s
1/2 (0.005)(300)^2 = (0.005)(9.8)h
h=4592m ≈ 4590m
The bullet will rise 4590 m before falling
1/2 mv^2 = mgh
m=0.005 kg v=300 m/s
1/2 (0.005)(300)^2 = (0.005)(9.8)h
h=4592m ≈ 4590m
The bullet will rise 4590 m before falling
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The problem is symmetrical. Up is the same as down
D = 0.5At^2
D = 0.5 * 9.8 m/s^2 * t^2
and
V=At
2 equations in 2,unknown
300 m/s = 9.8t
t = 300/9.8
Plug that in to the first equation and solve for D
D = 0.5At^2
D = 0.5 * 9.8 m/s^2 * t^2
and
V=At
2 equations in 2,unknown
300 m/s = 9.8t
t = 300/9.8
Plug that in to the first equation and solve for D
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Y = V²/2g = 300²/(2*9.8) = 4592 m
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V = AT, 300 m/s / 9.8 m/s/s = 30.61 seconds
displ = 1/2 AT^2 = .5 x 9.8 x 937.1 = 4592 meters
displ = 1/2 AT^2 = .5 x 9.8 x 937.1 = 4592 meters