part A,B,C,and D History Bookmarks Window Help History Book marks Window The Exp

Question

part A,B,C,and D History Bookmarks Window Help History Book marks Window The Expert TA I Human-like Grading, Automated ertta.com/Common/TakeTutorialAssignment.aspx il YouTube Wikipedia Popular C Q Google Bicycle Racer Sprints Near HW02 Begin Date: 9/10/2015 12:00:00 AM-Due Date: 9/14/2015 1159:00 PM End Date: 12/20/2013 11:59:00 PM a0% ) Problem 10: A quarter is dropped from a hot air balloon that is 295 m above the ground and nang at 8 5ml upward. problem use a coondinate system in which up is positive. 25% Part (a) Find the maxim um height tached for the coin in meters. 25% Part (b) Fnd its position, 400:after being released in m. 25% Part (c) Find its velocity 400) s fter being released ms. 25% Part (d) Find the time before it hits he ground in seconds. Grade Sammar PotentialI Submision sino l coso I tano ial (E171819 otanasingacos0 per atlemp) cos tanho cotanhO Degrees O Radiars llintsi deduction pei bial. Hals nmaning: Feedback: deductioe per feedback Submivgn History Feefback 8

Explanation / Answer

Initial height, hi = 295 m
initial vi = 8.5 m/s
a = -9.8 m/s^2
a)
let it travel d m upward to reach max height.
then:
vf^2 = vi^2 + 2*a*d
0 = 8.5^2 + 2* (-9.8) * d
d = 3.7 m
Maximum height = 295 m + 3.7 m = 298.7 m

b)
Find time taken to reach max height:
Vf = Vi + a*t
0 = 8.5 - 9.8 *t
t = 0.87 s
so basically need to calculate the distance moved by it after 4-0.87 = 3.13 s after reaching top
d = vi*t + 0.5*a*t^2
=0 *3.13 - 0.5*9.8*3.13^2
= -48 m
Its position is 298.7 - 48 = 250.7 m from ground

c) Basically need to calculate the Velocity after 3.13 s after reaching top
Vf = Vi + a*t
= 0 -9.8*3.13
= -30.7 m/s
Its moving downward with speed 30.7 m/s

d)
Time taken to reach bottom from max height be t'
d = 0.5*a*t^2
-298.7 = 0.5*(-9.8)*t'^2
t' = 7.8 s
Total time taken after it was dropped = 7.8 + 0.87 s = 8.67 s
Answer: 8.67 s

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.