Question

I have a Kinematics question I’m a bit confused about.It says:A cannon fires a shell straight upward; 1.5 s after it is launched, the shell is moving upward with a speed of 18 m/s. Assuming air resistance is negligible, find the speed (magnitude of velocity) of the shell at launch and 5.4 s after the launch.

Answer 1

Vf = v0 - at

Where:

V0= initial speed

a = acceleration = -g (gravity)

t= time

Vf= v0 - gt

18= 0 - (9.8 x 1.5 )

18 = vo - 14.7

18+14.7 = vo

V0= 32.7 m/s

Velocity of the shell at launch and 5.4 s after launch:

Vf = v0- gt

v = 32.7 - (9.8x 5.4)

v = 32.7-52.92

v= -20.22 m/s