16. 6.0 seconds after launch, the space shuttle is 529.2 m above the ground. a.
ID: 1330613 • Letter: 1
Question
16. 6.0 seconds after launch, the space shuttle is 529.2 m above the ground.
a. What is the space shuttle's acceleration?b What is the space shuttle's velocity after 3.0 seconds?
c What is the space shuttle's velocity at 6.0 seconds d What is the space shuttle's average velocity ater the first 6.0 seconds? e How high is the space shuttle after 3.0 seconds?
17. Melissa threw a penny straight down off the Empire State building. The building is 354 m tall. If Melissa threw the penny down such that it left her hand at 35 m/s, a. How fast will the coin be traveling when it hits the pavement?
b. How long will the coin be in the air?
18. An hour later, after the sidewalk damage was cleaned up, Paul dropped a coin off the top of the Empire State building.
a. How fast will the coin be traveling when it hits the pavement?
b. How long will the coin be in the air?
19. A methanol powered dragster travels a 1/4 mile from a stand still. The final speed of the best dragster will reach 300 mph.
a. Convert all units to standard SI units
b. Assuming the dragster’s acceleration to be constant, what will it be?
c. How long will the dragster take to finish the 1/4 mile?
Explanation / Answer
16.
Vi = initial velocity = 0 m/s
a = acceleration
d = distance travelled = 529.2 m
t = time taken = 6 sec
a)
Using the equation
d = Vi t + (0.5) a t2
529.2 = 0 t + (0.5) a (6)2
a = 29.4 m/s2
b)
Vf = velocity after 3 sec
t = 3 sec
using the equation
Vf = Vi + at
Vf = 0 + 29.4 x 3
Vf = 88.2 m/s
c)
Vf = velocity after 6 sec
t = 6 sec
using the equation
Vf = Vi + at
Vf = 0 + 29.4 x 6
Vf = 176.4 m/s
17.
Vi = initial velocity = 35 m/s
a = 9.8
d = 354 m
Using the equation
Vf2 = Vi2 + 2 ad
Vf2 = 352 + 2 x 9.8 x 354
Vf = 90.4 m/s
b)
Using the equation
Vf = Vi + at
90.4 = 35 + 9.8 t
t = 5.7 sec
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.