Question

A gun with a muzzle velocity of 65.2 m/s is fired horizontally from a tower. Neglecting air resistance, how far downrange in meters will the bullet be 2.7 seconds later

Answer 1

s ≈ 35.8 m

Step-by-step explanation:

Let's apply the uniform acceleration equation to find the vertical height travelled after 2.7 seconds.

`s = ut + (1)/(2)at^2`

s = distance travelled

u = initial vertical velocity

t = time taken

a = gravitational acceleration

The gun is fired horizontally, hence there is no initial vertical velocity.

`s = 0(2.7) + (1)/(2)(9.81)(2.7)^2`

`s = 35.75745 m`

s ≈ 35.8 m