Madeline fires a bullet horizontally. The rifle is 1.9 meters above the ground. The bullet travels 200 meters horizontally before it hits the ground. What speed did Madeline's bullet have when it exited the rifle?

321.027 m/s

Explanation:

First let´s analyze the vertical problem:

Always when an object is above the ground and nothing is holding it (like the bullet after being fired) the gravitational force will start acting on it.

Then the vertical acceleration of the bullet will be the gravitational acceleration, we can write this as:

a(t) = -g

Where g = 9.8m/s^2

and the negative sign is because this acceleration pulls the bullet downwards.

To get the vertical velocity we need to integrate over time, this will lead to:

v(t) = (-9.8m/s^2)*t + v0

where v0 is the initial vertical velocity, as the bullet is fired horizontally, there is no initial vertical velocity, then we have v0 = 0m/s

And the velocity equation is:

v(t) = (-9.8m/s^2)*t

Now for the vertical position, we need to integrate again, to get:

p(t)  =(1/2)*(-9.8m/s^2)*t^2 + p0

Where p0 is the initial vertical position, in this case, is 1.9 meters above the ground, then p0 = 1.9m

And the vertical position equation will be:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + 1.9 m

Now we want to find the time such that the vertical position is equal to zero, this will mean that the bullet it the ground.

p(t) = 0m = (1/2)*(-9.8m/s^2)*t^2 + 1.9 m

    (1/2)*(9.8m/s^2)*t^2 = 1.9m

    (4.9m/s^2)*t^2 = 1.9m

                  t = √(1.9m/(4.9m/s^2)) = 0.623 seconds.

This means that the bullet will travel for 0.623 seconds before hitting the ground.

Now we also can ignore the air friction for the horizontal motion, then we can assume that the horizontal speed does not change.

Then we can use the relationship:

Distance = speed*time

We know that:

time = 0.623 seconds

distance = 200m

now we can replace that in the equation to find the horizontal speed.

200m = speed*0.623s

200m/0.623s = speed = 321.027 m/s

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