Question

A bullet is shot vertically upward with an initial velocity of 128 ft/s. The bullet height after t seconds is y(t) = 128t - 16t². What is the

velocity when the bullet hits the ground?

Answer 1

We have the position function,
`y(t)=128t-16t^2`, that represents the height of a bullet shot from a rifle, we are also told that the initial velocity of the bullet is
`v_(0) =128 ft/s`.

To start, we need to find the time the bullet hits the ground. The bullet will hit the ground when it's position is equal to zero. So let
`y(t)=0` and solve for t using algebraic means.

`\Longrightarrow \hspace 00=128t-16t^2 \Longrightarrow \hspace00=-16t(t-8)`

So,
`t=0s` or
`t=8s`.

Logically the value of 8s would make sense as the bullet has to travel upwards and downwards. So it takes the bullet 8 seconds to hit the ground.

The question ask to find the velocity of the bullet when it hits the ground, we know time, now we just need to find the velocity function. We know that the derivative of the position function is equal to the velocity function (
`y'(t)=v(t)`). So take the derivative of
`y(t)`.

`\Longrightarrow y'(t)=v(t)=128-32t`

We need to know the velocity of the bullet as it hits the ground. We know the time at which the bullet hits the ground, so plug that value into our function for velocity.

`\Longrightarrow v(8)=128-32(8)\Longrightarrow v(8)=128-256 \Longrightarrow v(t)=-128ft/s`

Thus the velocity of the bullet right before it hits the ground is -128 ft/s.