Two identical bullets are used. Both are released at the same height - one fired out of a gun, the other is dropped. Ignoring air resistance, which hits the ground first?

Both bullets will hit the ground at the same time.

Explanation:

Let's only analyze the vertical problem.

Any object that is not in the floor or resting in some site is being affected by the gravitational force (remember that we are ignoring air resistance)

Then the acceleration of this object will be equal to the gravitational acceleration:

a = -9.8m/s^2

Where the minus sign is because this acceleration goes down.

To get the velocity equation we need to integrate over time, we will get:

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

Where v0 is the initial vertical velocity.

To get the position equation we need to integrate over time again, we will get:

p(t) = (1/2)*( -9.8m/s^2)*t^2 + v0*t + H

Where H is the initial height.

p(t) = (-4.9 m/s^2)*t^2 + v0*t + H

The object will hit the ground when p(t) = 0

Then we need to solve for t the next equation:

(-4.9 m/s^2)*t^2 + v0*t + H = 0

Notice that the only things we need to know are:

H = initial height (we know that is the same for both bullets)

v0 = initial vertical velocity (also is the same for both bullets)

Notice that the horizontal velocity does not affect this equation, then we will get the same value of t for the dropped bullet and for the fired bullet.

This means that both bullets will hit the ground at the same time.

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg