QuestionFind the ball's velocity when it first reaches $880 \mathrm{ft}$ , given $s(t)=-16 t^{2}+256 t$ .

Studdy Solution

STEP 1

Assumptions1. The position function of the ball is given by $s(t)=-16 t^{}+256 t$ . The initial velocity of the ball is $256 \, \mathrm{ft/s}$
3. The ball reaches a vertical height of $880 \, \mathrm{ft}$

STEP 2

First, we need to find the time $t$ when the ball reaches a height of $880 \, \mathrm{ft}$ . We can do this by setting the position function $s(t)$ equal to $880 \, \mathrm{ft}$ and solving for $t$ .
$-16 t^{2}+256 t =880$

STEP 3

Rearrange the equation to solve for $t$ .
$16 t^{2}-256 t +880 =0$

STEP 4

Divide the entire equation by16 to simplify.
$t^{2}-16 t +55 =0$

STEP 5

This is a quadratic equation in the form of $ax^2 + bx + c =0$ . We can solve for $t$ using the quadratic formula, $t = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$ .
$t = \frac{16 \pm \sqrt{16^2 -4155}}{2*1}$

STEP 6

Calculate the value inside the square root.
$\sqrt{16^2 -4155} = \sqrt{256 -220} = \sqrt{36} =6$

STEP 7

Substitute the value of square root in the formula and solve for $t$ .
$t = \frac{16 \pm6}{2}$

STEP 8

There are two possible values for $t$ , $t =11$ and $t =5$ . Since we are interested in the first time the ball reaches the height of $880 \, \mathrm{ft}$ , we take the smaller value, $t =5$ .

STEP 9

The velocity function $v(t)$ is the derivative of the position function $s(t)$ . So, we need to differentiate $s(t)$ to get $v(t)$ .
$v(t) = \frac{ds(t)}{dt} = -32t +256$

STEP 10

Substitute $t =5$ into the velocity function to find the velocity of the ball when it first reaches a height of $880 \, \mathrm{ft}$ .
$v(5) = -32*5 +256$

STEP 11

Calculate the velocity.
$v(5) = -32*5 +256 =96 \, \mathrm{ft/s}$ The velocity of the ball when it first reaches a vertical height of $880 \, \mathrm{ft}$ is $96 \, \mathrm{ft/s}$ .

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