10. (EP) A toy rocket launches from the ground (starting from rest) and rises st
ID: 1878138 • Letter: 1
Question
10. (EP) A toy rocket launches from the ground (starting from rest) and rises straight up with an acceleration of 4.0 m/s2. The rocket maintains this acceleration until it runs out of fuel at an altitude of 530 m. After running out of fuel, the acceleration of the rocket is the accelcration duc to gravity, pointing straight downward. (a) What is the rocket's velocity when it runs out of fuel? (b) How long does it take the rocket to reach this point? (c) What is the maximum altitude reached by the rocket? Hint: the rocket is not in free fall when accelerating at 4.0 m/s2, but it is in free fall when it runs out of fuel.Explanation / Answer
(A)a=4m/s 2 at height of 530 m it runs out of fuel . Rocket starts from rest ,hence intial velocity u= 0 , we need to find final velocity at 530 m ,substituting in v 2 - u 2 = 2*a * s we have v 2= 2* 4* 530.v = 65.11m/s.
(B) time taken by the rocket to reach height of 530 m is caiculated s=u*t+(0.5)*a*t 2 ,since u=0 ,we get 530= 0.5*4*t 2 .
t = 16.27 s
( C)maximum altitude reached by the rocket is calculated using acceleration due to gravity g ,where intial velocity is 65.11 m/s and final velocity is v=0 . Substituting in formula. v 2 - u 2 = 2*g*h since the rocket is falling down g= +g . We get h = (65.11)2/ 2* 9.8 = 216.3 m .
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.