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larisa86 [58]
3 years ago
14

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 86.0 m/s2

for 1.70 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?
Physics
1 answer:
Harman [31]3 years ago
5 0
<span>When the fuel  of the rocket is consumed, the acceleration would be zero. However, at this phase the rocket would still be going up until all the forces of gravity would dominate and change the direction of the rocket. We need to calculate two distances, one from the ground until the point where the fuel is consumed and from that point to the point where the gravity would change the direction. 

Given:
a = 86 m/s^2 
t = 1.7 s

Solution:

d = vi (t) + 0.5 (a) (t^2) 
d = (0) (1.7) + 0.5 (86) (1.7)^2 
d = 124.27 m 

vf = vi + at 
vf = 0 + 86 (1.7) 
vf = 146.2 m/s (velocity when the fuel is consumed completely) 

Then, we calculate the time it takes until it reaches the maximum height.
vf = vi + at 
0 = 146.2 + (-9.8) (t) 
t = 14.92 s

Then, the second distance
d= vi (t) + 0.5 (a) (t^2) 
d = 146.2 (14.92) + 0.5 (-9.8) (14.92^2) 
d = 1090.53  m

Then, we determine the maximum altitude:
 d1 + d2 = 124.27 m + 1090.53 m = 1214.8 m</span>
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Use the equation of motion to answer the question. Use the equation of motion to answer the question.
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The final position of the object after 2 s is 11 m.

Motion: This can be defined as the change in position of a body.

⇒ Formula:

  • x = x₀+v₀t+1/2(at²)........................ Equation 1

⇒ Where:

  • x = Final position of the object
  • x₀ = Starting position
  • v₀ = Starting velocity
  • t = time
  • a = acceleration

From the question,

⇒ Given:

  • x₀ = 4.5 m/s
  • t = 2 s
  • x₀ = 2m
  • a = 0 m/s²

⇒ Substitute these values into equation 1

  • x = 2+(4.5×2)+1/2(0²×2)
  • x = 2+9+0
  • x = 11 m

Hence, The final position of the object after 2 s is 11 m

Learn more about motion here: brainly.com/question/15531840

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