Projectile Motion and Quadratic Regression

Question

A quarterback throws a football from a height of 6 feet with an initial velocity of 20 feet per second. The height h of the ball at time t second can be represented by the function h(t)= -16t^2 +20t + 6. If the ball is caught at 2 feet, find the range if the function.

Answer 1

In the answer above, the max height is when v(t) = h'(t) means the derivative of the function which shows the rate of change of h(t), which is a calculus topic.

If you haven't covered that in your Algebra 2 class, picture what the equation h(t) = H(t) = -16t^2 + 20t + 6 looks like. The leading term of the quadratic equation is negative so H(t) is a downward facing parabola. The max height is called the vertex. If you put H(t) into the vertex form, y = a(x - h)^2 + k, the vertex, (h, k) or ( -b/(2a), H(-b/(2a) ) is the highest point.

From H(t) we get a = -; b = 20 and c = 6. h = -b/(2a) = -(20)/((2)(-16)) = + 20/32 = 5/8. Plug in 5/8 to H(t) and you get k = H(5/8) = -16(5/8)(5/8) + 20(5/8) + 6 = -25/4 + 50/4 + 24/4 = (-25 + 50 + 24)/4 = 49/4.

Answer 2

h = -16t^2 +20t + 6 = 2

-16t^2 +20t +4 = 0

-4t^2 +5t +1 = 0

t = 5/8 + (1/8)sqr(25+16) = 5/8 +(sqr39)/3 seconds when caught, when h=2 feet

range of function is from 0 to max height

max height is when v(t)=h'(t) = 0

-32t +20 = 0

t = 20/32 = 5/8

h(5/8) = max height = -16(5/8)^2 +20(5/8) + 6 =-25/4+25/2+6= 49/4= 12.25 feet

h ranges from 0 to 12 1/4 ft

or maybe they want you to say from 2 feet to 12 1/4, as if caught at 2 feet, it's never less than 2 feet high

[2, 12.25]= the range