Answer:
The ball throw into the air with an initial velocity of 22 meters per second is 27.4 meters above the ground after 3 seconds.
Step-by-step explanation:
The quadratic function [tex]h(t) = -4.9t^2 + 22t + 5.5[/tex] represents the height of the ball above the ground, h(t), in meters, with respect to time, t, in seconds.
To find h(3), substitute t = 3 into the function expression:
[tex]h(3)=-4.9\cdot 3^2+22\cdot 3+5.5\\ \\=-4.9\cdot 9+66+5.5\\ \\=-44.1+71.5\\ \\=27.4[/tex]
Meaning: the ball throw into the air with an initial velocity of 22 meters per second is 27.4 meters above the ground after 3 seconds.
