Question

During halftime of a soccer ​game, a sling shot launches​ T-shirts at the crowd. A​ T-shirt is launched from a height of 4 feet with an initial upward velocity of 64 feet per second. The​ T-shirt is caught 39 feet above the field. How long will it take the​ T-shirt to reach its maximum​ height? What is the maximum​ height? What is the range of the function that models the height of the​ T-shirt over​ time?

Answer 1

Answer:

Step-by-step explanation:

We are dealing with the "american" measure of feet, so the pull of gravity that we need is -16t^2.  The quadratic that models this, in standard form, is

$s(t)=-16t^2+v_{0}t+h_{0}$

where v0 is the initial vertical velocity (64) and h0 is the initial height (4).  Filling that standard position function in with our stuff gives us:

$s(t)=-16t^2+64t+4$

We are asked how long it will take for the object to reach its max height, then of course we are asked what the max height is (of course!).  I am going to use calculus here since it's the easiest way to go.  The first derivative of the position function is the velocity function.  

v(t) = -32t + 64

You should know at this point from either algebra 2 or physics or calculus that an object has no velocity at its max height because it has to stop in the air in order to turn around and head back down.  So we will set the velocity function equal to 0 and solve for how long it takes to reach that max height.

0 = -32(t - 2) so

t = 2 seconds, which answers the first question.  If it takes 2 seconds to reach its max height, we can sub in 2 for t in the position function to find the object's position at 2 seconds.  This will be the max height.

$s(2)=-16(2)^2+64(2)+4$ so

s(2) = 68 feet.  The max height is 68 feet, which answers the seconds question.

As far as the range goes, the position function is an upside down parabola with a max height of 68.  So the range, if I'm understanding what is being asked, is y | y ≤ 68