Question

Tyrone launches a toy rocket into the air with an initial upward velocity of 49 ft/s and an initial height of 0 ft. a. How long will it take the rocket to reach its maximum height? Round to the nearest hundredth. b. How high above the ground will it be? Round to the nearest tenth. c. What is the range of the function? d. Why is there a maximum point on the graph of this function?

Answer 1

Step-by-step explanation:

You're using ft/s here, so i had to convert acceleration into ft/s^2 which is 32.174

a) At its maximum height, v=0 so we use the equation of motion

v = u + at

0 = 49 + (-32.174)t

t = 49 / 32.174

t = 1.52 seconds (2 dp)

b) Using this time:

s = 1/2 (u + v)t

s = 1/2 (49 +0)1.52

s = (49 x 1.52)/2

s = 37.3 feet (1dp)

c) The range is 0<=y<=37.3

d)

There is a limit as as the rocket goes off, it speeds down due to friction, then as it slows down to a halt (at v=0), it turns around and begins to slide. The turnaround point is the limit for the charts.