Question

A cannon fires a shell straight upward; 2.3 s after it is launched, the shell is moving upward with a speed of 17 m/s. Assuming air resistance is negligible, find the speed (magnitude of velocity) in meters per second of the shell at launch and 5.4 s after the launch.

Answer 1

Answer:

The speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.                    

Explanation:

Let u is the initial speed of the launch. Using first equation of motion as :

$u=v-at$

a=-g

$u=v+gt\\\\u=17+9.8\times 2.3\\\\u=39.54\ m/s$

The velocity of the shell at launch and 5.4 s after the launch is given by :

$v=u-gt\\\\v=39.54-9.8\times 5.4\\\\v=-13.38\ m/s$

So, the speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.