Question

a cannonball is fired at a 45.0 degree angle and an initial velocity of 625 m/s assume no air resistance.
how long does it take for the cannonball to hit the ground?
how high did the cannonball travel?
how far did the cannonball travel horizontally when it hit the ground

Answer 1

Take g = 9.81 m/s^2 and split the launch velocity into components:
v0x = v0 cos45° = v0/√2, v0y = v0 sin45° = v0/√2.

With v0 = 625 m/s:
v0x = v0y = 625/√2 ≈ 441.94 m/s.

1) Time of flight:
T = 2 v0y / g = 2(441.94)/9.81 ≈ 90.1 s.

2) Maximum height:
hmax = v0y^2 /(2g) = (441.94^2)/(2·9.81) = (390625·0.5)/(19.62) ≈ 9.95×10^3 m ≈ 9955 m.

3) Horizontal range:
R = v0^2 sin(2·45°)/g = v0^2 / g = 625^2 / 9.81 ≈ 3.98×10^4 m ≈ 39.8 km.

Answers (rounded): T ≈ 90.1 s, hmax ≈ 9.95×10^3 m, R ≈ 3.98×10^4 m.