Math

QuestionA ball is thrown from a 112 ft building with an initial velocity of 96 ft/s. Find when it hits the ground and passes the building.

Studdy Solution

STEP 1

Assumptions1. The initial height of the ball is112 feet. . The initial velocity of the ball is96 feet per second.
3. The acceleration due to gravity is -32 feet per second squared.
4. The distance ss (in feet) of the ball from the ground after tt seconds is given by the equation s(t) =112 +96t -16t^.
5. The ball strikes the ground when s(t)=0s(t) =0.
6. The ball passes the top of the building on its way down when the velocity is negative.

STEP 2

First, we need to find the time when the ball strikes the ground. We can do this by setting s(t)=0s(t) =0 and solving for tt.
0=112+96t16t20 =112 +96t -16t^2

STEP 3

Rearrange the equation to be in standard quadratic form at2+bt+c=0at^2 + bt + c =0.
16t296t+112=016t^2 -96t +112 =0

STEP 4

Now, we can solve for tt using the quadratic formula t=b±b24ac2at = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}.
t=96±(96)2416112216t = \frac{96 \pm \sqrt{(-96)^2 -4*16*112}}{2*16}

STEP 5

Calculate the discriminant b24acb^2 -4ac.
(96)2416112=9217168=2048(-96)^2 -4*16*112 =921 -7168 =2048

STEP 6

Substitute the discriminant back into the quadratic formula to find the values of tt.
t=96±204832t = \frac{96 \pm \sqrt{2048}}{32}

STEP 7

Calculate the two possible values for tt.
t=96+20432,9620432t = \frac{96 + \sqrt{204}}{32}, \frac{96 - \sqrt{204}}{32}

STEP 8

Calculate the actual values for tt.
t=96+45.254832,9645.254832=4.4,1.4t = \frac{96 +45.2548}{32}, \frac{96 -45.2548}{32} =4.4,1.4

STEP 9

Since time cannot be negative, the ball strikes the ground after4.4 seconds.

STEP 10

Next, we need to find the time when the ball passes the top of the building on its way down. We can do this by finding when the velocity is negative. The velocity v(t)v(t) is given by the derivative of s(t)s(t).
v(t)=dsdt=9632tv(t) = \frac{ds}{dt} =96 -32t

STEP 11

Set v(t)=0v(t) =0 and solve for tt.
0=9632t0 =96 -32t

STEP 12

olve for tt.
t=9632=t = \frac{96}{32} =So, the ball passes the top of the building on its way down after seconds.

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