What is the speed of the rocket when it is 350 above the surface of the earth?
A rocket starts from rest and moves upward from the surface of the earth. For the first 10 of its motion, the vertical acceleration of the rocket is given by a(y)=2.9, where the -direction y+ is upward.
To find the speed of the rocket when it is 350 meters above the surface of the earth, we need to utilize the kinematic equations.
Given:
Initial velocity (u) = 0 m/s (rocket starts from rest)
Vertical acceleration (a(y)) = 2.9 m/s^2 (upward direction)
Vertical displacement (s) = 350 m
We can use the second equation of motion, which relates the final velocity (v), initial velocity (u), acceleration (a), and displacement (s):
v^2 = u^2 + 2as
Substituting the given values into the equation:
v^2 = 0^2 + 2 * 2.9 * 350
v^2 = 0 + 2030
v^2 = 2030
Taking the square root of both sides:
v = √2030
v ≈ 45.05 m/s
Therefore, the speed of the rocket when it is 350 meters above the surface of the earth is approximately 45.05 m/s.
To determine the speed of the rocket when it is 350 m above the surface of the Earth, we can use the equations of motion and the given information about the acceleration.
We know that the rocket starts from rest, so its initial velocity (v₀) is 0 m/s.
The equation of motion in the y-direction is given by:
v = v₀ + at
where:
v is the final velocity,
v₀ is the initial velocity (0 m/s),
a is the acceleration (2.9 m/s²), and
t is the time.
First, we need to find the time it takes for the rocket to reach the height of 350 m. We can use the following kinematic equation:
s = s₀ + v₀t + 0.5at²
where:
s is the displacement from the surface of the Earth (350 m),
s₀ is the initial displacement (0 m),
v₀ is the initial velocity (0 m/s),
a is the acceleration (2.9 m/s²), and
t is the time.
Plugging in the values, we have:
350 = 0 + 0 + 0.5(2.9)t²
Simplifying the equation, we get:
350 = 1.45t²
Dividing both sides by 1.45, we have:
t² ≈ 241.38
Taking the square root of both sides, we find:
t ≈ √241.38
t ≈ 15.53 seconds (rounded to two decimal places)
Now that we know the time it takes for the rocket to reach 350 m above the surface of the Earth, we can find its final velocity by substituting the values into the equation of motion:
v = v₀ + at
v = 0 + 2.9(15.53)
v ≈ 45.00 m/s (rounded to two decimal places)
Therefore, the speed of the rocket when it is 350 m above the surface of the Earth is approximately 45.00 m/s.