A bullet is shot from a height of 5 meters with a velocity of 230 m/s. How long before the bullet hits the ground? How far will it travel?

Question

A bullet is shot from a height of 5 meters with a velocity of 230 m/s. How long before the bullet hits the ground? How far will it travel? 

Answer 1

Solution Attached:

https://i.imgur.com/Kbqctui.jpeg

Answer 2

The bullet follows a parabola resulting from a combination of two movements:

- A horizontal trajectory due to inertia, which is what the bullet would

follow in the absence of gravity.

- A vertical fall due to gravity, which is what the bullet would follow

in the absence of any initial horizontal velocity.

You can view the trajectory in an X-Y coordinate space, where

the X coordinate is along the ground, and

the Y coordinate is flush against the muzzle of the gun.

Then [x(t), y(t)] is the position of the bullet after time t.

(We start measuring time t the instant the bullet leaves the gun.)

x(t) = v.t

y(t) = h - g.t.t/2,

where

v = 230m/s is the initial horizontal velocity

h = 5 m is the height of the gun

g = 10 m/s^2 is approximately the gravitational acceleration

The fall continues until the bullet hits the ground at some time t1,

which we can calculate because at time t1, y(t1) = 0. That is,

h - g.t1.t1/2 = 0

t1 = sqrt(2h/g) = sqrt(10m/(10m/s^2)) = sqrt(1s^2) = 1s

Having obtained t1, we can calculate how far the object got horizontally by:

x(t1) = v.t1 = 230m/s . 1s = 230m