A sounding rocket, launched vertically upward with an initial speed of
70.0 m/s,
accelerates away from the launch pad at
5.00 m/s2.
The rocket exhausts its fuel, and its engine shuts down at an altitude of
1.30 km,
after which it moves freely under the influence of gravity.
(a) How long is the rocket in the air?
(b) What is the maximum altitude reached by the rocket?
(c) What is the velocity of the rocket just before it strikes the ground?
To find the answers to these questions, we can break the problem down into different stages of the rocket's motion.
(a) How long is the rocket in the air?
To find the total time the rocket is in the air, we need to first calculate the time it takes for the rocket to reach its maximum altitude, then add the time it takes to fall back to the ground.
1. Calculate the time it takes for the rocket to reach its maximum altitude:
The initial speed of the rocket is 70.0 m/s, and it accelerates at 5.00 m/s^2. We can use the kinematic equation:
v = u + at
where:
v = final velocity (in this case, it will be 0 m/s at the maximum altitude)
u = initial velocity (70.0 m/s)
a = acceleration (5.00 m/s^2)
t = time
Rearranging the equation, we get:
t = (v - u) / a
Substituting the values:
t = (0 - 70.0) / 5.00
Calculating:
t = -70.0 / 5.00
t = -14.0 seconds
Since time cannot be negative, we discard the negative sign. Therefore, the time taken for the rocket to reach its maximum altitude is 14.0 seconds.
2. Calculate the time it takes for the rocket to fall back down to the ground:
To find the time it takes for the rocket to fall back down, we use the equation for motion under constant acceleration again. This time, the initial velocity is 0 m/s, and the acceleration is -9.81 m/s^2 (due to gravity). The final position is the altitude from which the rocket was launched, which is 1.30 km.
Using the kinematic equation:
s = ut + (1/2)at^2
Where:
s = final position (1.30 km = 1300 m)
u = initial velocity (0 m/s)
a = acceleration (-9.81 m/s^2)
t = time
Rearranging the equation, we get a quadratic equation in terms of t:
(1/2)at^2 + ut - s = 0
Plugging in the values, we have:
(1/2)(-9.81)t^2 + 0t - 1300 = 0
Simplifying the equation, we get:
-4.905t^2 - 1300 = 0
Solving for t using the quadratic formula, we find two possible values for t: t = 50.7s and t = -50.7s. We discard the negative value since time cannot be negative, so the time it takes for the rocket to fall back down to the ground is 50.7 seconds.
3. Calculate the total time the rocket is in the air:
The total time the rocket is in the air is the sum of the time taken to reach the maximum altitude (14.0 seconds) and the time it takes to fall back down (50.7 seconds).
Total time = 14.0 + 50.7
Total time = 64.7 seconds
Therefore, the rocket is in the air for approximately 64.7 seconds.
(b) What is the maximum altitude reached by the rocket?
To find the maximum altitude reached by the rocket, we need to calculate the displacement (change in position) when the rocket reaches its maximum altitude.
Using the same kinematic equation as before:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity (70.0 m/s)
a = acceleration (5.00 m/s^2)
t = time (14.0 seconds)
Plugging in the values:
s = (70.0)(14.0) + (1/2)(5.00)(14.0)^2
Calculating:
s = 980 + (1/2)(5.00)(196)
s = 980 + 490
s = 1470 m
Therefore, the maximum altitude reached by the rocket is 1470 meters (or 1.47 km).
(c) What is the velocity of the rocket just before it strikes the ground?
The velocity of the rocket just before it strikes the ground can be found using the equation:
v = u + at
Where:
v = final velocity
u = initial velocity (0 m/s since the rocket has come to a stop at its maximum altitude)
a = acceleration due to gravity (-9.81 m/s^2, acting downwards)
t = time (50.7 seconds)
Plugging in the values:
v = 0 + (-9.81)(50.7)
Calculating:
v = -498.87 m/s
Since velocity cannot be negative in this context (as the rocket is moving downwards), we discard the negative sign.
Therefore, the velocity of the rocket just before striking the ground is approximately 498.87 m/s.