Question

A cannonball is fired at a 45.0° angle and an initial velocity of 670 m/s. Assume no air resistance. How long until the cannonball hits the ground?
96.7 s
55.3 s
65.3 s
45.5 s

Answer 1

Answer:

96.7 s

Explanation:

Time of flight in projectile can be calculated thus:

T = 2 × u × sin ϴ/ g

Where;

T = time of flight (s)

u = initial velocity (m/s)

ϴ = Angle of projectile (°)

g = acceleration due to gravity (9.8m/s²)

Based on the provided information; u = 670m/s, ϴ = 45°

Hence, using T = 2.u.sin ϴ/ g

T = 2 × 670 × sin 45° ÷ 9.8

T = 1340 × 0.7071 ÷ 9.8

T = 947.52 ÷ 9.8

T = 96.68

T = 96.7s