Question

6. A cannon shoots a ball at 50 m/s at an angle of 53.13°. Ignore the original height of the cannon
and assume level ground.
a. How long is the cannon ball in the air?
50 m/s
53.13°
​

Answer 1

Answer:

(a) For the first part

Y = V0yt -0.5gt2 since the cannonball hits the ground Y=0

0 = 50*sin30*t - 0.5*9.8*t2 because V0y = 50sin30

0 = 25*t -4.9*t2

0 = (25 - 4.9t)t

t = 0 before firing the cannon and t =25/4.9 = 5.1 seconds when the cannonball hit the ground

(b) ΔX is the difference between the range of the cannonball and the calculated horizontal distance

R = V02 sin2(θ)/g = 502*sin60/9.8 =220.92 m

X = V0x * t = 50cos30*5.1 = 220.84 m

ΔX = R - X =220.92 - 220.84 = 0.10 m

ΔX = 10 cm