Cannon Ball Question?

a) use d=1/2gt^2 where d=height and t is the time to reach the height, and g is the acceleration of gravity:

=> t=(2*d/g)^0.5 = (2*(4507.55m)/(9.8m/s^2))^0.5

= 30.33 seconds to reach the apex. It should take the same time to hit the ground again, so the total time in the air is 2 x that:

time in the air = 61 sec

b) assume the force of 9,600,000N acts on the ball for the time that the ball is in the cannon barrel so:

F=ma => a=F/m = (9,600,000N/120kg) = 80,000 m/s^2

v=a*t = (80,000m/s^2)*(0.005s) = 400m/s

We also know:

time to reach apex = (vertical velocity vy)/g

=> vertical velocity = (time to reach apex)*(g)

= (30.33sec)*(9.8m/s^2) = 297 m/s

So we can calculate theta:

sin(theta) = vy/v = (297m/s)/(400m/s) = 0.7425

=> theta = 47.9 degrees

c) vx = v*cos(theta) = (400m/s)*(0.6700) = 268m/s

d= vx*t = (268m/s)*(60.66s) = 16,300m

d) d=1/2at^2 = 0.5*(80,000m/s^2)*(0.005s)^2

= 1 m

Hope this helps.