Question

A cannon fires a cannon ball at an angle of θ0 = 30.0° above the horizontal axis. The initial speed of the ball is 395 m/s . Ignore air resistance for all calculations.

A) Find the maximum height h that the ball attains.

B) Find the horizontal distance (range) R the cannon ball would travel given the above (the same as for Part A) conditions.

Answer 1

Split up vertical and horizontal speed into two separate and independent things, and use y velocity to calculate the time until it hits the ground, and that time combined with the x velocity for the distance.

v0 = 395 m/s
30 degree angle -> vx = cos(30)*v0 = 342.08 ; vy = sin(30)*v0 = 197.5
h = vy*t - (1/2)g*t²
vy = vy(begin) - g*t
You want to know the highest point, where vy = 0
vy(begin) - g*t = 0
v0*sin(30) = g*t
t = v0*sin(30)/g = 197.5/9.81
t = 20.13 s
for the height of that time:
h = vy*t - (1/2)g*t²
h = v0*sin(30)*20.13 - (1/2)*9.81*(20.13)²
h = 1988.09 m

You want to find out the time until it hits the ground, so
vy*t - (1/2)g*t² = 0
v0*sin(30)*t - (1/2)*g*t^2 = 0
t(v0*sin(30) - (1/2)*g*t) = 0
-> ball is on the ground at either t = 0 (obvious, but not helpful) or v0*sin(30) - (1/2)*g*t = 0
v0*sin(30) - (1/2)*g*t = 0
v0*sin(30) = (1/2)*g*t
2vy*sin(30) = g*t
t = (2*v0*sin(30))/g
t = (2*395*sin(30))/9.81
t = 40.265 s
Now you use that time with vx to find out how far it flies
s = vx*t
s = v0*cos(30)*t
s = 395*cos(30)*40.265 =13773.85 meters