Question

A bullet of mass 0.037 kg is fired horizontally with a speed of 77 m/s from a tall bridge. If the bullet is in the air for 2.1 s. How far from the base of the bridge does it land in m?

Answer 1

Starting from the assumption that for the time traveled, the bullet has no loss in its speed and considering the cinematic equations of movement description we have to

`v = (x)/(t)`

Where,

x = Displacement

t = Time

Replacing with our values and rearranging to find the displacement we have

`x = v*t\\x = 11*2.15\\x = 161.7m`

The distance from the base of the bridge where it lands is 161.7m