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QuestionFind when the ball hits the ground and when it passes the building using S(t)=64+48t−16t2S(t)=64+48t-16t^2.

Studdy Solution

STEP 1

Assumptions1. The equation of motion for the ball is (t)=64+48t-16t^, where (t)(t) is the height of the ball at time tt. . The ball strikes the ground when (t)=0(t) =0.
3. The ball reaches the top of its trajectory when its velocity is zero, which is given by the derivative of (t)(t).

STEP 2

To find when the ball strikes the ground, we need to solve the equation (t)=0(t) =0 for tt.0=64+48t−16t20 =64 +48t -16t^2

STEP 3

Rearrange the equation to make it easier to solve.16t2−48t+64=016t^2 -48t +64 =0

STEP 4

This is a quadratic equation in the form ax2+bx+c=0ax^2 + bx + c =0. We can solve it using the quadratic formula t=−b±b2−4ac2at = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}.

STEP 5

Substitute a=16a =16, b=−48b = -48, and c=64c =64 into the quadratic formula.
t=−(−48)±(−48)2−4∗16∗642∗16t = \frac{-(-48) \pm \sqrt{(-48)^2 -4*16*64}}{2*16}

STEP 6

implify the equation.
t=48±2304−409632t = \frac{48 \pm \sqrt{2304 -4096}}{32}

STEP 7

Calculate the value under the square root.
t=48±−179232t = \frac{48 \pm \sqrt{-1792}}{32}

STEP 8

The value under the square root is negative, which means there are no real solutions. This is a mistake. The equation was incorrectly rearranged in3. The correct rearrangement is16t2−48t+64=016t^2 -48t +64 =0to16t2−48t−64=016t^2 -48t -64 =0

STEP 9

Substitute a=16a =16, b=−48b = -48, and c=−64c = -64 into the quadratic formula.
t=−(−48)±(−48)2−4∗16∗(−64)2∗16t = \frac{-(-48) \pm \sqrt{(-48)^2 -4*16*(-64)}}{2*16}

STEP 10

implify the equation.
t=48±2304+409632t = \frac{48 \pm \sqrt{2304 +4096}}{32}

STEP 11

Calculate the value under the square root.
t=48±640032t = \frac{48 \pm \sqrt{6400}}{32}

STEP 12

implify the square root.
t=48±8032t = \frac{48 \pm80}{32}

STEP 13

Calculate the two possible values for tt.
t=48+8032=t = \frac{48 +80}{32} =t=48−8032=−t = \frac{48 -80}{32} = -

STEP 14

Since time cannot be negative, we discard the negative solution. The ball strikes the ground after4 seconds.

STEP 15

To find when the ball reaches the top of its trajectory, we need to find when the velocity is zero. The velocity is given by the derivative of (t)(t), ′(t)'(t).
′(t)=48−32t'(t) =48 -32t

STEP 16

Set the derivative equal to zero and solve for tt.
0=48−32t0 =48 -32t

STEP 17

Rearrange the equation to solve for tt.
32t=4832t =48

STEP 18

Divide both sides by32 to solve for tt.
t=4832t = \frac{48}{32}

STEP 19

implify the fraction to find the time when the ball reaches the top of its trajectory.
t=1.5t =1.5The ball passes the top of the building on its way down after1.5 seconds.

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