Question

A model rocket was launched from a podium 5 meters above are

an initial velocity of 98 m/s. The function that models height in me

with respect to time (in seconds) is h(t) = 5 + 987-4.9t2

meters above ground at

meters)

Answer 1

See solution below

Explanation:

Find the complete question below;

Given the height modeled by the function;

h(t) = 5 + 98t-4.9t²

when t = 0

h(0) = 5 + 98(0)-4.9(0)²

h(0) = 5

when t = 5

h(5) = 5 + 98(5)-4.9(5)²

h(5) = 5 + 490 - 122.5

h(5) = 372.5

when t = 10

h(10) = 5 + 98(10)-4.9(10)²

h(10) = 5 + 980 - 490

h(10) = 495

when t = 15

h(15) = 5 + 98(15)-4.9(15)²

h(15) = 5 + 1470 - 1,102.5

h(15) = 372.5

when t = 20

h(10) = 5 + 98(20)-4.9(20)²

h(10) = 5 + 1960 - 4.9(400)

h(10) = 1965- 1960

h(10) = 5

Find the graph attached