A quarterback throws a football to a teammate. The football is 6.5ft above the ground when it leaves the quarterback's hand. His teammate catches it 3.5s later, at a height above the ground of 5 ft. Which quadratic model

y = - 16t² + 55.6t + 6

Step-by-step explanation:

Using y - y₀ = vt - 1/2gt² where g = 32 ft/s², and v the velocity of the football

So y = y₀ + vt - 1/2 × (32 ft/s²)t²

y = y₀ + vt - 16t² where y₀ = 6.5 ft

y = 6 + vt - 16t²

Now, when t = 3.5 s, that is the time the teammate catches the ball after the quarterback throws it, y = 5 ft. Substituting these into the equation, we have

5 = 6.5 + v(3.5 s) - 16(3.5 s)²

5 = 6.5 + 3.5v - 196

collecting like terms, we have

5 - 6.5 + 196 = 3.5v

194.5 = 3.5v

v = 194.5/3.5 = 55.57 ft/s ≅ 55.6 ft/s

So, substituting v into y, our quadratic model is

y = 6 + 55.6t - 16t²

re-arranging, we have

y = - 16t² + 55.6t + 6

x = central angle apc = measure of minor arc ac

the two angles x and 30 add to 180, so x+30 = 180 and x = 180-30 = 150. the measure of minor arc ac is 150 degrees.

what we need to do from here is find the circumference and then multiply it by x/360 since there are 360 degrees in a circle. in this case, x = 150, so x/360 = 150/360 = 15/36 = 5/12. the shortest distance along the curve from a to c is 5/12 of the distance around the entire circle (aka circumference)

c = circumference

c = pi*d

c = pi*24

c = 24pi

now multiply this by 5/12 to get

distance along minor arc ac = (5/12)*(circumference)

distance along minor arc ac = (5/12)*(24pi)

distance along minor arc ac = 10pi

the formula you can use is

l = (x/360)*pi*d

with

l = length of minor arc

x = central angle

d = diameter

which essentially is the steps we took though a bit differently.