A model rocket is launched straight upward
with an initial speed of 45.4 m/s. It accelerates with a constant upward acceleration of
2.98 m/s
2
until its engines stop at an altitude
of 150 m.
What is the maximum height reached by
the rocket? The acceleration of gravity is
9.81 m/s
2
When does the rocket reach maximum height?
Answer in units of s
How long is the rocket in the air?
Answer in units of s
See your 10-2-11,11:32am post for solution.
To find the maximum height reached by the rocket, we can use the kinematic equation:
vf^2 = vi^2 + 2aΔy
Where:
vf = final velocity (0 m/s when the rocket reaches maximum height)
vi = initial velocity (45.4 m/s)
a = acceleration (2.98 m/s^2)
Δy = change in height (unknown)
Rearranging the equation, we get:
Δy = (vf^2 - vi^2) / (2a)
As we're looking for the maximum height, we can substitute vf = 0:
Δy = (-vi^2) / (2a)
Substituting the known values:
Δy = (-45.4^2) / (2 * 2.98)
= -2065.16 / 5.96
≈ -346.70 m
The result is a negative value because the direction of motion is upwards, and we set the initial height as 0. So, the maximum height reached by the rocket is approximately 346.70 meters above the initial height.
To find when the rocket reaches maximum height, we can use the equation:
vf = vi + at
Rearranging the equation for time:
t = (vf - vi) / a
Substituting the given values:
t = (0 - 45.4) / 2.98
= -45.4 / 2.98
≈ -15.23 s
Since time cannot be negative, the rocket reaches the maximum height at approximately 15.23 seconds.
To calculate how long the rocket is in the air, we can use the equation:
vf = vi + at
Rearranging the equation for time:
t = (vf - vi) / a
Substituting the given values:
t = (0 - 45.4) / (-9.81)
= -45.4 / (-9.81)
≈ 4.63 s
Again, time cannot be negative, so the rocket is in the air for approximately 4.63 seconds.