A model rocket was launchee from a podium 5 meters above groubd at a initial velocity of 98 m/s. The function that models height (in meters) with respect to time (in seconds) is h(t)=5+98t-4.9t²

Given : A model rocket was launchee from a podium 5 meters above groubd at a initial velocity of 98 m/s. The function that models height (in meters) with respect to time (in seconds) is h(t)=5+98t-4.9t²

Refer the attached figure below for the complete question.

Solution :

Part A: Function h(t)=5+98t-4.9t²

When t=0,

[tex]h(0)=5+98(0)-4.9(0)^2[/tex]

[tex]h(0)=5[/tex]

When t=5,

[tex]h(5)=5+98(5)-4.9(5)^2[/tex]

[tex]h(5)=372.5[/tex]

When t=10,

[tex]h(10)=5+98(10)-4.9(10)^2[/tex]

[tex]h(10)=495[/tex]

When t=15,

[tex]h(15)=5+98(15)-4.9(15)^2[/tex]

[tex]h(15)=372.5[/tex]

When t=20,

[tex]h(20)=5+98(20)-4.9(20)^2[/tex]

[tex]h(20)=5[/tex]

Complete table form is

Time (seconds) 0 5 10 15 20

Elevation (meters) 5 372.5 495 372.5 5

Part B :

Plot the points from table and then connect them with a smooth curve to graph h(t),

Refer the graph below.