Question

A cannon fires a shell straight upward; 1.8 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 5.5 s after the launch

Answer 1

To solve this problem we will apply the concepts of acceleration according to the laws of kinetics. Acceleration can be defined as the change in velocity per unit time, that is,

$a = \frac{v-u}{t}$

Here,

v = Final velocity

u = Initial velocity

t = Time

Rearranging to find the initial velocity,

$u = v-at$

Now the acceleration is equal to the gravity, then,

$u = v+gt$

$u = 17+(9.8)(1.8)$

$u = 34.64m/s$

The velocity of the shell at 5.5s after the launch is,

$v = u+at$

$v = u-gt$

$v = 34.64-(9.8)(5.5)$

$v = -19.26m/s$

Therefore the magnitude of the velocity of the shell at 5.5s is -19.26m/s