Question

In: Physics

A rocket, initially at rest on the ground, accelerates upward with a constant acceleration of 94.0...

A rocket, initially at rest on the ground, accelerates upward with a constant acceleration of 94.0 m/s2 until it reaches a speed of 1.50×102 m/s when the engines are cut off. After that the rocket is in free-fall.
What is the maximum height reached by the rocket ?

What total time elapses between take-off and the rocket

hitting the ground?

Solutions

Expert Solution

This problem can be divided into three different processes.

1. When the rocket is accelerated upwards from rest to the velocity of 1.5*102 m/s.

The initial velocity of the rocket u is 0.

The final velocity of the rocket v= 1.5*102 m/s

Acceleration due to the engine is 94.0 m/s2 upwards.

Acceleration due to gravity is 9.8 m/s2 downwards.

Total acceleration a during this process is (94.0-9.8) m/s2 = 84.2 m/s2

height attained during this process can be determined by the equation of motion: v2 = u2 + 2as

Substituting the above values to determine the height s1 reached in this process,

(1.5*102 m/s)2 = 02 + (2*84.2 m/s2 *s1)

22500/(168.4) = s1

s1= 133.6 m

Time required to attain this height can be determined by the equation: v=u+at

1.5*102 m/s = 0+ 84.2*t1

t1= 1.8 s

2. When the engines are cut off, the rocket continues to go upwards upto a certain height before it stops.

In this process, the initial velocity once the engines are cut off is u=1.5*102 m/s

The rocket continues to travel upwards to a height s2, where the velocity becomes zero, i.e. v=0 m/s

The only acceleration acting on the rocket during this process is the acceleration due to gravity a= 9.8 m/s, which is acting downwards.

Again, by using the equation of motion v2 = u2 + 2as, we determine s2.

0=(1.5*102)2 + (2*(-9.8)* s2)

s2= (1.5*102)2 /19.6

s2= 1147.6 m

Time required to attain this height in this process can be determined by v=u+at

0=(1.5*102 )- 9.8t2

t2= (1.5*102 )/9.8

t2= 15.3 s

Thus, the maximum height reached by the rocket is s1+s2=133.6+1147.6 m =1281.2 m

3. The rocket then falls back to the ground in free-fall

We know that the initial velocity at the beginning of the fall is u=0

Only acceleration acting on it is gravitational, a=9.8 m/s

Height from which the fall takes place is s= 1281.2 m

We can determine the time required to for the fall by s=ut+ (1/2)at2

1281.2 = 0 + (1/2)*9.8* t32

1281.2*2/9.8 = t32

t3= 16.17 s

Thus the total time for the rocket to take the flight from take off to hitting the ground is t1+t2+t3

= 1.8+15.3+16.17 s

Total time = 33.27 s

genius_generous answered 2 years ago
Next > < Previous

Related Solutions

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that...
A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 91.3 m/s2 for 1.79 seconds, at which point its fuel abruptly runs out. Air resistance has no effect on its flight. What maximum altitude (above the ground) will the rocket reach?
A test rocket is launched, starting on the ground, from rest, by accelerating it along an incline with constant acceleration "a". The incline ha
A test rocket is launched, starting on the ground, from rest, by accelerating it along an incline with constant acceleration "a". The incline has length "L", and it rises at ? degrees above the horizontal. At the instant the rocket leaves the incline, its engines turn off and it is subject only to gravity, g?+9.81m/s2.  (Air resistance can be ignored). Taking the usual x-y coordinate system, with an origin at the top edge of the incline,  (a)what is the position vector when the rocket is at its highest point?  (b)What is the position vector when the rocket is on its way back down and once again at the same height as the top edge of the incline?  Your symbolic answer should only depend on a, L,?, g, and/or numerical factors Asked by
a toy rocket is fired fired at rest vertically upwards and accelerates at a constant rate....
a toy rocket is fired fired at rest vertically upwards and accelerates at a constant rate. At the instant the engine cuts out, the rocket has risen to 52 m and acquired a velocity of 80 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground with negligible air resistance. The speed of the rocket upon impact on the ground is closest to
A rocket is fired straight upward, starting from rest with an acceleration of 25.0 m/s2. It...
A rocket is fired straight upward, starting from rest with an acceleration of 25.0 m/s2. It runs out of fuel at the end of 4.00 s and continues to coast upward, reaching a maximum height before falling back to Earth. (a) Find the rocket’s height when it runs out of fuel; (b) find the rocket’s velocity when it runs out of fuel; (c) find the maximum height the rocket reaches; (d) find the rocket’s velocity the instant before the rocket...
A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds...
A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is given by a(t) = 60t ft/s2 , at which time the fuel is exhausted and it becomes a freely “falling” body and falls to the ground. (a) Determine the position function s(t) for all times t > 0. (b) What is the maximum height achieved by the rocket? ( c) At what value of t does the rocket land?
A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds...
A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a(t)=96t, at which time the fuel is exhausted and it becomes a freely "falling" body. 1919 seconds later, the rocket's parachute opens, and the (downward) velocity slows linearly to −16 ft/s in 5 s. The rocket then "floats" to the ground at that rate. (a) Determine the position function s and the velocity function v(for all times t). v(t)=    if 0≤t≤3    ...
A 7400 kg rocket blasts off vertically from the launch pad with a constant upward acceleration...
A 7400 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.15 m/s2 and feels no appreciable air resistance. When it has reached a height of 550 m , its engines suddenly fail so that the only force acting on it is now gravity. a)What is the maximum height this rocket will reach above the launch pad? b)How much time after engine failure will elapse before the rocket comes crashing down to the launch...
A 7800 kg rocket blasts off vertically from the launch pad with a constant upward acceleration...
A 7800 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.15 m/s2 and feels no appreciable air resistance. When it has reached a height of 575 m , its engines suddenly fail so that the only force acting on it is now gravity. A) What is the maximum height this rocket will reach above the launch pad? b)How much time after engine failure will elapse before the rocket comes crashing down to the...
A 7450 kg rocket blasts off vertically from the launch pad with a constant upward acceleration...
A 7450 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25 m/s2 and feels no appreciable air resistance. When it has reached a height of 500 m , its engines suddenly fail so that the only force acting on it is now gravity. A. What is the maximum height this rocket will reach above the launch pad? B. How much time after engine failure will elapse before the rocket comes crashing down to...
an elevator starts from rest and moves upward with a constant acceleration 1.2 abolt of mass...
an elevator starts from rest and moves upward with a constant acceleration 1.2 abolt of mass m in the elevator celling above the elevator nworks loose and falls dowen assume g = 9.8 what is the velocity of the bolt at is it hit the flor relative to the ground? what is the velocity of the bolt as it hits the floor relative to the elevator
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT