Asked by Dennis
A three-kilogram cannonball is shot out of a cannon with an initial velocity of 300 m/s at a 25o angle. A headwind exerts a constant 5 N horizontal force. How far will the cannonball travel before horizontally hitting the ground?
Answers
Answered by
Anonymous
vertical problem first, how long in air?
v = Vi - g t
v = 300 sin 25 - 9.81 t
at top v = 0 (half the flight)
9.81 * t at top = 300 * sin 25
t at top = 12.9 seconds upward
so
total t = 25.8 seconds aloft
Now horizontal
a = F/m = -5/3
v = 300 cos 25 - (5/3) t
x = 300 cos 25 t - (5/6) t^2
put in t = 25.8 and you have it
v = Vi - g t
v = 300 sin 25 - 9.81 t
at top v = 0 (half the flight)
9.81 * t at top = 300 * sin 25
t at top = 12.9 seconds upward
so
total t = 25.8 seconds aloft
Now horizontal
a = F/m = -5/3
v = 300 cos 25 - (5/3) t
x = 300 cos 25 t - (5/6) t^2
put in t = 25.8 and you have it
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