A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 79.0 m/s at ground level. The engines then fire, and the rocket accelerates upward at 3.90 m/s2 until it reaches an altitude of 980 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of −9.80 m/s2. (You will need to consider the motion while the engine is operating and the free-fall motion separately.)
(a) For what time interval is the rocket in motion above the ground?
[40.51 seconds]
(b) What is its maximum altitude in kilometers?
[? km]
(c) What is its velocity just before it hits the ground?
[? m/s]
V^2 = Vo^2 + 2a*h
V^2 = 79^2 + 7.8*980 = 13,885
V = 117.83 m/s. @ 980 m
a. V = Vo + a*t = 117.83
79 + 3.9t = 117.83
3.9t = 117.83-79 = 38.83
Tr1 = 9.96 s. to rise to 980 m.
V = Vo + g*t
117.83 - 9.8t = 0
9.8t = 117.83
Tr2 = 12.02 s to reach max. ht.
hmax = ho + (V^2-Vo^2)/2g
hmax=980 + (0-(117.83)^2)/-19.6=1688 m.
= max. ht
hmax = Vo*t + 0.5g*t^2 = 1688 m
0 + 4.9t^2 = 1688
t^2 = 345
Tf = 18.57 s. = Fall time.
T = Tr1 + Tr2 + Tf
T = 9.96+12.02+18.57 = 40.55 Seconds.
b. hmax = 1688 m. = 1.688 km. See previous calculation.
c. V^2 = Vo^2 + 2g*hmax
V^2 = 0 + 19.6*1688 = 33,084.8
V = 182 m/s.
To solve this problem, we need to analyze the motion of the rocket in two different phases: when the engine is operating and when it is in free fall.
(a) For what time interval is the rocket in motion above the ground?
Let's find the time it takes for the rocket to reach an altitude of 980 m while the engine is operating. We can use the kinematic equation:
h = h0 + v0t + (1/2)at^2
where:
h is the altitude (980 m),
h0 is the initial altitude (0 m),
v0 is the initial velocity (79.0 m/s),
a is the acceleration (3.90 m/s^2), and
t is the time.
Rearranging the equation and substituting the known values, we get:
980 = 0 + (79.0)t + (1/2)(3.90)t^2
Simplifying the equation, we have:
1.95t^2 + 79t - 980 = 0
Using the quadratic formula, we find:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = 1.95, b = 79, and c = -980.
Plugging in the values, we get:
t = (-79 ± √(79^2 - 4(1.95)(-980))) / (2(1.95))
Solving for t, we find two possible values for t: t1 = 8.81 s and t2 = -47.21 s.
Since time cannot be negative in this context, we can discard t2. Therefore, the rocket takes approximately 8.81 seconds to reach an altitude of 980 m while the engine is operating.
Next, let's find the time it takes for the rocket to hit the ground in free fall. We can use the equation:
h = h0 + v0t + (1/2)at^2
where:
h is the altitude (0 m),
h0 is the initial altitude (980 m),
v0 is the initial velocity (unknown), but we can assume it is the velocity just before the engines fail,
a is the acceleration due to gravity (-9.80 m/s^2), and
t is the time.
Rearranging the equation, we have:
0 = 980 + v0t + (1/2)(-9.80)t^2
Simplifying the equation, we get:
-4.9t^2 + v0t + 980 = 0
Since the rocket is in free fall, its initial velocity just before it hits the ground is 0 (v0 = 0).
Solving the quadratic equation again, we find:
t = (-v0 ± √(v0^2 - 4(-4.9)(980))) / (2(-4.9))
Simplifying, we get:
t = √(v0^2 + 980) / 4.9
Since this is the time it takes for the rocket to hit the ground, we can substitute this expression into the equation we found earlier:
8.81 = √(v0^2 + 980) / 4.9
Squaring both sides of the equation, we have:
77.6161 = (v0^2 + 980) / 24.01
Multiplying both sides by 24.01, we get:
1863.3 = v0^2 + 980
Rearranging the equation, we find:
v0^2 = 1863.3 - 980
v0^2 = 883.3
Taking the square root of both sides, we have:
v0 ≈ ±29.74 m/s
Since the velocity cannot be negative in this context, the velocity just before the rocket hits the ground is approximately 29.74 m/s.
Therefore, the time interval for which the rocket is in motion above the ground is approximately 8.81 seconds.
Let's break down the problem into different phases of motion and solve it step by step.
Phase 1: Rocket's initial acceleration upward by the catapult.
The rocket is launched with an initial speed of 79.0 m/s upward. We need to find the time it takes for the rocket to reach an altitude of 980 m while accelerating at 3.90 m/s^2.
Step 1: Find the time taken to reach the altitude of 980 m.
We can use the kinematic equation:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity
a = acceleration
t = time taken
Rearranging the equation to solve for time (t) gives us:
t = (v - u) / a
Using:
s = 980 m
u = 79.0 m/s
a = 3.90 m/s^2
Substituting the values into the equation, we get:
t = (980 - 79.0) / 3.90
t ≈ 231.28 seconds
So, the time taken for the rocket to reach an altitude of 980 m is 231.28 seconds. However, we need to consider only the time interval when the rocket is in motion above the ground, so we subtract the time it takes for the initial acceleration:
t_interval = t - (2 * [(v - u) / a])
Substituting the values, we get:
t_interval ≈ 231.28 - [(2 * (79 - 0) / 3.90]
t_interval ≈ 231.28 - [162 / 3.90]
t_interval ≈ 231.28 - 41.54
t_interval ≈ 189.74 seconds
So, the time interval for which the rocket is in motion above the ground is approximately 189.74 seconds.
Answer to (a): The rocket is in motion above the ground for approximately 189.74 seconds.
Phase 2: Rocket's free-fall motion.
After the engine fails, the rocket goes into free fall, with an acceleration of -9.80 m/s^2.
Step 2: Find the maximum altitude reached by the rocket.
Using the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement
At the maximum altitude, the final velocity is 0 m/s. Rearranging the equation, we get:
s = (v^2 - u^2) / (2a)
Using:
v = 0 m/s
u = 79.0 m/s
a = -9.80 m/s^2
Substituting the values into the equation, we get:
s = (0^2 - 79.0^2) / (2 * -9.80)
s ≈ (-6241) / (-19.6)
s ≈ 318.78 meters
To convert the maximum altitude to kilometers, we divide by 1000:
s_km = 318.78 / 1000
s_km ≈ 0.31878 kilometers
So, the maximum altitude reached by the rocket is approximately 0.31878 kilometers.
Answer to (b): The maximum altitude of the rocket is approximately 0.31878 kilometers.
Phase 3: Rocket's velocity just before it hits the ground.
During free fall, we can find the velocity using the equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time taken
We need to find the time it takes for the rocket to hit the ground from the maximum altitude.
Step 3: Find the time taken to hit the ground.
Using the equation:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity
a = acceleration
t = time taken
At the maximum altitude, the final displacement is 0 m. Rearranging the equation, we get:
t = √(2s / a)
Using:
s = 318.78 m (maximum altitude)
a = -9.80 m/s^2
Substituting the values into the equation, we get:
t = √(2 * 318.78 / -9.80)
t ≈ √(-650.56)
t is imaginary (negative square root)
Since we can't have a negative time in this context, we discard this solution.
Answer to (c): The rocket will not hit the ground (fails to reach the ground).
In conclusion:
(a) The rocket is in motion above the ground for approximately 189.74 seconds.
(b) The maximum altitude reached by the rocket is approximately 0.31878 kilometers.
(c) The rocket does not hit the ground.