Answer:
The maximum height that a cannonball fired at 420 m/s at a 53.0° angles is 5740.48m.
hmax = 5740.48 m
Explanation:
This is an example of parabolic launch. A cannonball is fired on flat ground at 420 m/s at a 53.0° angle and we have to calculate the maximum height that it reach.
V₀ = 420m/s and θ₀ = 53.0°
So, when the cannonball is fired it has horizontal and vertical components:
V₀ₓ = V₀ cos θ₀ = (420m/s)(cos 53°) = 252.76 m/s
V₀y = V₀ cos θ₀ = (420m/s)(cos 53°) = 335.43m/s
When the cannoball reach the maximum height the vertical velocity component is zero, that happens in a tₐ time:
Vy = V₀y - g tₐ = 0
tₐ = V₀y/g
tₐ = (335.43m/s)/(9.8m/s²) = 34.23s
Then, the maximum height is reached in the instant tₐ = 34.23s:
h = V₀y tₐ - 1/2g tₐ²
hmax = (335.43m/s)(34.23s)-1/2(9.8m/s²)(34.23s)²
hmax = 11481.77m - 5741.29m
hmax = 5740.48m
