Calculus analytical

Question

During the first 40 s of a rocket flight, the rocket is pro pelled straight up so that in t seconds it reaches a height of s= 0.313 ft.

(a) How high does the rocket travel in 40 s? (b) What is the average velocity of the rocket during the first 40 s?

(c) What is the average velocity of the rocket during the

first 1000 ft of its flight? (d) What is the instantaneous velocity of the rocket at the end of 40 s?

Answer 1

h(40) = -16(40)^2 +vo(40) +ho = .313

ho=0

h(40) = -16(1600) +40vo = .313

40vo = .313+ 25600= 25600.313

vo = 25600.313/40 = 640.007825= initial velocity at time t=0

h(40) = -16(40)2 +640.007825(40) = -25600+25600.313

= 0.313 feet in 40 seconds

= about 3.756 inches

h'(t) = -32t +640.007825 = 0

t = 640.007825/32 = about 20 seconds to reach maximum height

in 40 seconds it's virtually hitting the ground

max h(t) = h(20) = -16(20)^2 + 640.007825(20) =

about 12800-6400= 6400 feet

in about 20 seconds it travels up about 6,400 feet

in about 40 seconds it travels up 6400 ft, then down 6400 feet for a total distance traveled of 12800 feet

initially, instantaneous velocity = 640.007825 ft/sec

in about 20 seconds velocity = 0

in about 40 seconds instantaneous velocity =-640.007825 ft/sec

that's all with time t rounded off to nearest integer for seconds