Question

(1 point) A student with a third floor dormitory window 32 feet off the ground tosses a water balloon straight up in the air with an initial velocity of 16 feet per second. It turns out that the instantaneous velocity of the water balloon is given by the velocity function v(t)=−32t+16, where v is measured in feet per second and t is measured in seconds. Let s(t) represent the height of the water balloon above the ground at time t, and note that s is an antiderivative of v. That is, v is the derivative of s: s′(t)=v(t). Find a formula for s(t) that satisfies the initial condition that the balloon is tossed from 32 feet above ground. In other words, make your formula for s satisfy

Answer 1

s(t) = -16t^2 + 16t + 32 (feet)

Explanation:

The time function for s(t) [the height of the balloon] is the integral with respect to time [t] of the velocity function, v(t) = -32t + 16. This integral is

t^2

s(t) = -32------ + 16t + s0, where t is time and so is the initial height of the

2 balloon.

Thus we have in this particular case:

t^2

s(t) = -32------ + 16t + 32, where t is time and so is the initial height of the

2 balloon.

We can simplify this to:

s(t) = -16t^2 + 16t + 32, where t is time in seconds and s is in feet.

Answer 2

`s(t) = -16t^(2) + 16t + 32`

Explanation:

The position s(t) is the integrative of the velocity.

The velocity is given by the following equation:

`v(t) = -32t + 16`

So the position is:

`s(t) = \int {(-32t + 16)} \, dt`

`s(t) = -16t^(2) + 16t + C`

In which c is the initial position, that is 32 feet above ground, so +32.

The equation is:

`s(t) = -16t^(2) + 16t + 32`