Arocket is launched vertically from the ground with an initial velocity of 64 ft/sec . (a) write a quadratic function h(t) that shows the height, in feet, of the rocket t seconds after it was launched. (b) graph h(t) on the coordinate plane. (c) use your graph from part 3(b) to determine the rocket’s maximum height, the amount of time it took to reach its maximum height, and the amount of time it was in the air. explain if you can. i'm really terrible at this part of math. make sure to: -write and graph the quadratic function. -determines the rocket’s maximum height, the amount of time it took to reach its maximum height, and the amount of time it was in the air.

answer:

a-

b - graph is shown below

c - maximum height is 19.401 meters, maximum time is 1.99 seconds and total time the rocket was in the air is 3.98 seconds.

step-by-step explanation:

we are given that,

the initial velocity of the rocket = 64 feet per second = 19.5 meter per second

part a: the equation of motion is given by , where

s= h(t) is the distance traveled above (in meters),

a= acceleration= -9.8 meter per second (as rocket is decelerating),

u= initial velocity (in meter per second),

t= time (in seconds).

on substituting the values,

i.e.

thus the quadratic equation is.

part b: the graph of the function is shown below.

part c: so, from the graph, we see that,

the maximum height of the rocket is 19.401 meters, which is obtained after 1.99 seconds.

as, the rocket was in the air for 1.99 seconds to reach the maximum.

and, it will take 1.99 seconds to return back to the ground.

hence, the total time the rocket was in the air is 1.99 + 1.99 = 3.98 seconds.

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