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 (in feet) of the ball from the ground after seconds is given by the equation s(t) =112 +96t -16t^.
5. The ball strikes the ground when .
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 and solving for .
STEP 3
Rearrange the equation to be in standard quadratic form .
STEP 4
Now, we can solve for using the quadratic formula .
STEP 5
Calculate the discriminant .
STEP 6
Substitute the discriminant back into the quadratic formula to find the values of .
STEP 7
Calculate the two possible values for .
STEP 8
Calculate the actual values for .
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 is given by the derivative of .
STEP 11
Set and solve for .
STEP 12
olve for .
So, the ball passes the top of the building on its way down after seconds.
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