u = 86 cos 32 = 72.9 m/s forever
Vi = 86 sin 32 = 45.6 up at start
x distance = u t
265 = (86 cos 32) t
t = 3.63 seconds in the air
h = Hi + Vi t - 4.9 t^2
h = 0 + 45.6(3.63) -4.9(3.63)^2
h = 101 meters high
v = Vi - 9.81 t
at t = 3.63
v = 10 m/s up
u = 72.9 still
speed = sqrt (100 + 72.9^2)
about 73.6 m/s
A cannon ball is fired from ground level with a speed of 86.0 m/s at an angle of 32 deg. above the horizontal. A 8.5 m tall wall is located 265 m away from the cannon. By how much does the cannon ball clear the wall?
In the previous problem, what is the speed of the cannon ball when it is directly above the wall?
1 answer
1 answer