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

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."