A bullet shot vertically into the air at +600 ft/s j. How high does the bullet rise above the ground?

Question

A bullet shot vertically into the air at +600 ft/s j. How high does the bullet rise above the ground? What is the velocity of the bullet when it returns to the ground? Assume air friction?

Answer 1

How high does the bullet rise above the ground?

The equation that relates the bullet's height (h) to its velocity (v) is

v² – (v₀)² = –2gh where v₀ = iinitial velocity given to be 600 ft/s.

At the top of it's trajectory the velocity of the bullet (v) is zero.

Plugging v = 0 into the above equation gives us –(v₀)² = –2gh_max

which means that (v₀)² = 2gh_max (where h_max is the maximum height, in feet).

We solve this equation for h_max by dividing both sides by 2g.

The maximum height of the bullet (h_max) is given by

h_max = (v₀)²/2g where v₀ = 600 ft/s and g = 32 ft/s²

Plugging in these values into the last equation gives us

h_max = (600)²/(2 × 32)

What is the velocity of the bullet when it returns to the ground?

The bullet's velocity when it returns to the ground will be v = 600 ft/s, the same as it's initial velocity.

A way to think about this is that the bullet starts out with just kinetic energy (related to its velocity (v) by the the equation KE = ½mv²).

As it goes upward all of it's kinetic energy is converted to the same amount of potential energy (related to its height (h) by the the equation PE = mgh).

As the bullet comes back down all of its potential energy is converted back to the same amount of kinetic energy that it started with. The same amount of kinetic energy means the same velocity.

Assume air friction?

No. Unless it is stated otherwise air resistance is ignored.