Question

A cannonball is shot from a cannon at a launch angle of 63 and an initial velocity of 147 m/s. The gravitational acceleration is -9.8 m/s^2 . Assuming NO air resistance, how long the cannonball remain in the air.

A.)40.2s
B.)13.4 s
C.)26.7 s
D.)9.8 s

Answer 1

Vy = V sin theta = 147 * sin 63 = 131 m/s

Time to reach zero vertical speed tu= 131 m/s / 9.8 m/s^2 = 13.4 sec

Time for ball to fall to original speed

td = 131 m/s 9.8 m/s^2 = 13.4 sec

Total time in air = tu + td = 13.4 s + 13.4 s = 26.8 s

Or 2 * sin 63 * 147 / 9.8 = 26.7 sec