Answer:
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
Answer:
y
=
−
x
+
9
Step-by-step explanation:
Plz Mark brainliest!
Unfortunately, there is no given table. However, assuming no air resistance, the projectile moves in constant acceleration in the y-axis, while constant velocity in the x-axis. If the total time of flight is 6 seconds, then at 3 seconds the reaches its maximum height. Also, there is no given initial velocity, which means any useful calculation is impossible.
Do you mean checking your work? In that case it would be 9*20.
