Quadratic Functions

Question

During halftime of a football 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 47 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

When thinking about this problem we have to make some assumptions first. We need to assume we are using Earth’s gravity of 9.8m/s^2 or 32.2ft/s^2. We also need to assume that air resistance can be ignored. We also need to know that the equation for the vertical position of an object over time is:

y0 + v0*t - 1/2 * g * t^2

y0 is the initial position which is 4ft

v0 is the initial velocity which is 64ft/s

g is acceleration due to gravity which we’ll say is 32ft/s^2

So from what we know, our equation will look something like:

y = 4 + 64t - 16t^2

y is the height off the ground.

t is the time in seconds.

Using the Quadratic function we get a=-16, b=64 and c=4.

Note: sqrt( ) is the square root function

t = (-b + sqrt(b^2 - 4*a*c))/(2*a) and

t = (-b - sqrt(b^2 - 4*a*c))/(2*a)

t = (-64 + sqrt(64^2 - 4*(-16)(4)/(2*(-16)) and

t = (-64 - sqrt(64^2 - 4*(-16)(4)/(2*(-16))

t = -0.06

t = 4.06

That gives us y = (t + 0.06)(t - 4.06)

The find the highest point we need to find the vertex so we use the equation t = -b/2a where t is the time coordinate of the vertex and we can substitute that in to the original equation to find the y coordinate.

t = -64/2*(-16)

t = 2

y = 4 + 64(2) - 16(2)^2

y = 68

So after 2 seconds we will have reached our maximum height of 68ft and started to fall.

If we update our quadratic equations to replace y = 0 with our ending point y = 47 we get:

47 = 4 + 64t - 16t^2

0 = -43 + 64t - 16t^2

a=-16, b=64 and c=-43

Applying the quadratic formula again we get:

t = (-64 + sqrt(64^2 - 4*(-16)(-43))2*(-16)) and

t = (-64 - sqrt(64^2 - 4*(-16)(-43))(2*(-16))

t = 0.854356

t = 3.14564

So if you were thinking about it carefully you might have already expected to see that the t-shirt would be at a height of 47ft at two different times.

So going back to the original questions. The maximum height could be 47ft at time 0.85 seconds never reaching the maximum possible height, or the maximum height is 68ft at time 2 seconds follow by it reaching 47ft again at 3.15 seconds. The range for the physical t-shirt would be [4,68], mathematically it would be (-inf,68]. The domain would be [0, 3.15], mathematically (-inf,inf).