Chrome File Edit View History Bookmarks People Window Help PHYS 107 Cengage 9 Ho

Question

Chrome File Edit View History Bookmarks People Window Help PHYS 107 Cengage 9 How Do I Ask A Question On x CSecure https://www.webassign.net/web/Student/Assignment-Responses/submit?dep 18819464 QUESTION If, instead, some fuel remains, at what height should the engines be fired again to brake the rocket's fall and allow a perfectly soft landing? (Assume the same acceleration as in the initial descent.) PRACTICE IT Use the worked example above to help you solve this problem. A rocket moves straight upward, starting from rest with an acceleration of +26.5 m/s2. It runs out of fuel at the end of 5.01 s and continues to coast upward, reaching a maximum height before falling back to Earth. (a) Find the rocket's velocity and position at the end of 5.01 s m/s (b) Find the maximum height the rocket reaches. (c) Find the velocity the instant before the rocket crashes on the ground m/s EXERCISE HINTS: GETTING STARTED I P'M STUCK An experimental rocket designed to land upright falls freely from a height of 2.41 x 102 m, starting at rest. At a height of 96.6 m, the rocket's engines start and provide constant upward acceleration until the rocket lands. What acceleration is required if the speed on touchdown is to be zero? (Neglect air resistance.) m/s2 Need Help? LRead is

Explanation / Answer

Given,

a = 26.5 m/s^2 ; t = 5.01 s ;

a)We know that

S = ut + 1/2 at^2

S = 0 + 0.5 x 26.5 x 5.01^2 = 332.58 m

v = u + at

v = 0 + 26.5 x 5.01 = 132.77 m/s

Hence, vb = 132.77 m/s ; yb = 332.58 m

b)time taken

t = 132.77/9.81 = 13.53 s

h = 132.77 x 13.53 + 0.5 x -9.81 x 13.53^2 = 898.46 m

H = 898.46 + 332.58 = 1231.04 m

Hence, H = 1231.04 m

c)v = sqrt (2gh)

v = sqrt (2 x 9.81 x 1231.04) = 155.41 m/s

Hence, v = 155.41 m/s

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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