Question

For Questions 12-12, during halftime of a football game, a slingshot launches T-shirts at the crowd. A T-shirt is launched from a height of 3 feet with an initial upward velocity of 72 feet per second. The T-shirt is caught 50 feet above the field. What is the maximum height of the T-shirt? What is the range of the function?

Answer 1

What angle was the shot? I assume it was not shot straight up.
So, unless you provide the angle of launch, or the distance to the catcher, there is no way to pin down the max height or the range.

Given a launch angle of θ, you know the height at a horizontal distance x is

h = 3 + tanθ x - 16/(72 cosθ)^2 x^2
So, either you know x when h=50
or you need to know the angle θ

Then recall that the range is
R = 72^2/16 sin2θ

Answer 2

the relevant free fall equation is ... h = -16 t^2 + 72 t + 3

max height is on the axis of symmetry of the parabola (quadratic)
... tmax = -72 / (2 * -16)
... plug tmax back into the equation to find hmax

the range is from the ground to hmax

Answer 3

Nice catch, Scott. I neglected to read the word "upward."
And I also mistook the range of the function to be the range of the missile.

Answer 4

well from 3 feet to max h

Answer 5

Y = Yo + g*Tr = 0.
72 + (-32)Tr = 0,
Tr = 2.25 s. = Rise time.

a. Y^2 = Yo^2 + 2g*h = 0.
72^2 + (-64)h = 0,
h = 81 Ft. above launching point.
hmax = 81 + 3 = 84 Ft. above gnd.

h = 0.5g*Tf^2 = 50.
16Tf^2 = 50,
Tf = 1.77 s. = Fall time.

Range = Xo * (Tr+Tf) = Xo * (2.25+1.77).
Yo = Vo*sinA = 72.
Xo = Vo*CosA.

To find the range , I'll need to know the initial velocity(Vo) or the launch angle(A).