Question
A bronze cannon ball with a mass of 4.5 kg is fired vertically from a small cannon and reaches a height of 10.0 m. If the canon is now aimed at an angle 33 degrees above the horizontal, calculate the horizontal distance the cannon ball would travel before it reaches the ground.
Answers
Not much of a cannon :)
time to fall to ten meters = time to rise ten meters
10 = (1/2) g t^2 = 4.9 t^2
t = 1.43 seconds up or down
v = Vi - g t
0 = Vi -9.81 (1.43)
Vi = 14 m/s muzzle velocity
now at 33 deg
u = 14 cos 33 = 11.8 m/s forever
Vi = 14 sin 33 = 7.62 initial speed up
How long to top?
0 = 7.62 - 9.81 t
t = .777 seconds going up
so
time in air = 1.55 s
so horizontal distance = 11.8*1.55 =18.3 meters
time to fall to ten meters = time to rise ten meters
10 = (1/2) g t^2 = 4.9 t^2
t = 1.43 seconds up or down
v = Vi - g t
0 = Vi -9.81 (1.43)
Vi = 14 m/s muzzle velocity
now at 33 deg
u = 14 cos 33 = 11.8 m/s forever
Vi = 14 sin 33 = 7.62 initial speed up
How long to top?
0 = 7.62 - 9.81 t
t = .777 seconds going up
so
time in air = 1.55 s
so horizontal distance = 11.8*1.55 =18.3 meters
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