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 table and graph.

Step-by-step explanation:

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,

When t=5,

When t=10,

When t=15,

When t=20,

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.

answer:   28.56 feet (approx)

step-by-step explanation:

since, triangle abc has measure of angle c equal to 55 degrees, measure of angle abc equal to 90 degrees, and length of bc equal to 20 feet.

we have to find out the measurement of segment ab.

therefore, tan 55° =

but bc = 20 feet.

⇒ tan 55 ° =  

⇒ ab = 20 × tan 55 °

⇒ ab = 20 × 1.42814800674

⇒ ab = 28.5629601348   ≈ 28.56 feet