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A toy rocket is launched with an acceleration of 10.0 m/s2 upwards for 3.0 s. It then slows
down with an acceleration of 10.0 m/s2 in the opposite direction until it reaches its maximum
altitude. How high does it go?

Sagot :

MathPhys

Answer:

90 m

Explanation:

The rocket undergoes constant acceleration. We can use kinematic equations, also known as SUVAT equations, to model the rocket's motion. For this problem, we will use the following equations:

s = ut + ½ at²

v = u + at

v² = u² + 2as

where

  • s is the displacement
  • u is the initial velocity
  • v is the final velocity
  • a is the acceleration
  • t is the time

During the first part of the rocket's motion, it starts from rest (u = 0) and accelerates upward at a = 10.0 m/s² for a time of t = 3.0 s. The displacement after this time is:

s = ut + ½ at²

s = (0) (3.0) + ½ (10.0) (3.0)²

s = 45 m

The velocity of the rocket after this time is:

v = u + at

v = 0 + (10.0) (3.0)

v = 30.0 m/s

Next, the rocket decelerates from this speed (u = 30.0 m/s) at a rate of a = -10.0 m/s² until it reaches a stop (v = 0 m/s). The displacement of the rocket at this point is:

v² = u² + 2as

(0)² = (30.0)² + 2 (-10.0) s

s = 45 m

The rocket travels up 45 meters during acceleration, then another 45 meters during deceleration. The total height reached is:

45 m + 45 m = 90 m

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