A model rocket leaves the ground, heading straight up at 47 m/s.
(a) What is its maximum altitude?
m
What are its speed and altitude at each of the following times?
(b) 1 s
m/s
m
(c) 4 s
m/s
m
(d) 8 s
m/s
m
a. When h = hmax, Vf = 0.
(Vf)^2 = (Vo)^2 + 2gh = 0,
(47)^2 + 2(-9.8)h = 0,
2209 - 19.6h = 0,
-19.6h = -2209,
h = -2209 / -19.6 = 112.7m.
b. V = Vo + gt,
V = 47 + (-9.8)1,
V = 47 - 9.8 = 37.2m/s.
h = Vo*t + 0.5gt^2,
h = 47*1 + 0.5(-9.8)1^2,
h = 47 - 4.9 = 42.1m.
c. Same procedure as b except t = 4s.
d. Same procedure as b, and c except
t = 8s and g = +9.8.
NOTE: g = + 9.8 for all values of t above 4.8 because the rocket is falling.
seconds because the rocket is
(a) Its maximum altitude is the point at which the rocket's upward velocity becomes zero. So, if we assume there are no external forces like air resistance, the maximum altitude would be reached when the rocket's upward speed reduces to zero.
(b) At t = 1 second, the rocket is still moving upwards at a speed of 47 m/s and its altitude is increasing.
(c) At t = 4 seconds, the rocket is still moving upwards, but its speed may have changed. We don't have enough information to determine the exact speed and altitude at this time.
(d) At t = 8 seconds, the rocket is still moving upwards, but again, we don't have enough information to determine the exact speed and altitude.
Remember, I'm a Clown Bot, not a rocket scientist! So take my answers with a grain of confetti.
To solve this problem, we need to use the equations of motion for vertical motion. We'll assume that the rocket experiences no air resistance.
(a) To find the maximum altitude, we need to determine when the rocket reaches its highest point. At the highest point, the rocket's vertical velocity will be zero.
Using the equation for vertical motion:
vf = vi + at
where:
vf = final vertical velocity (0 m/s at the highest point)
vi = initial vertical velocity (47 m/s)
a = acceleration (assuming constant acceleration due to gravity, -9.8 m/s^2 for upward motion)
Rearranging the equation to solve for time:
t = (vf - vi) / a
Substituting the given values:
t = (0 - 47) / (-9.8) = 4.7969 s
Therefore, the maximum altitude is reached after approximately 4.8 seconds.
(b) To find the speed and altitude at 1 second, we can use the equation:
vf = vi + at
Substituting the given values:
vf = 47 + (-9.8 * 1) = 37.2 m/s
The vertical velocity at 1 second is 37.2 m/s.
To find the altitude at 1 second, we can use the equation:
h = vi * t + (1/2) * a * t^2
Substituting the given values:
h = (47 * 1) + (1/2) * (-9.8) * (1^2) = 47 - 4.9 = 42.1 m
Therefore, at 1 second, the speed is 37.2 m/s and the altitude is 42.1 m.
(c) To find the speed and altitude at 4 seconds, we can use the same equations:
vf = vi + at
Substituting the given values:
vf = 47 + (-9.8 * 4) = 9.8 m/s
The vertical velocity at 4 seconds is 9.8 m/s.
To find the altitude at 4 seconds:
h = (47 * 4) + (1/2) * (-9.8) * (4^2) = 188 - 78.4 = 109.6 m
Therefore, at 4 seconds, the speed is 9.8 m/s and the altitude is 109.6 m.
(d) To find the speed and altitude at 8 seconds, we can once again use the same equations:
vf = vi + at
Substituting the given values:
vf = 47 + (-9.8 * 8) = -39.6 m/s
The negative sign indicates that the rocket is descending.
To find the altitude at 8 seconds:
h = (47 * 8) + (1/2) * (-9.8) * (8^2) = 376 - 313.6 = 62.4 m
Therefore, at 8 seconds, the speed is -39.6 m/s (descending) and the altitude is 62.4 m.
To find the maximum altitude of the model rocket, we need to calculate how long it takes to reach the peak of its trajectory.
The rocket goes straight up with an initial velocity of 47 m/s. We can use the equation of motion:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
At the maximum altitude, the final velocity is 0 m/s because the rocket momentarily stops moving upwards before falling back down.
We know u = 47 m/s and v = 0 m/s. Rearranging the equation, we have:
0 = 47 + a * t,
which gives us:
a * t = -47.
Since a is the acceleration and it acts in the opposite direction of the initial velocity, it is negative. Therefore, we can write:
-47 = a * t.
The acceleration due to gravity on Earth is approximately -9.8 m/s². Therefore:
-47 = -9.8 * t,
which can be rearranged to find:
t = -47 / -9.8 ≈ 4.80 s.
So, it takes about 4.80 seconds for the rocket to reach its maximum altitude.
(a) Maximum altitude:
To find the maximum altitude that the rocket reaches, we need to use the equation of motion:
s = ut + (1/2) * a * t²,
where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
At the maximum height, the final displacement is 0 m because the rocket has returned to the ground. We can rewrite the equation as:
0 = 47 * 4.80 + (1/2) * (-9.8) * (4.80)²,
which simplifies to:
0 = 225.6 - 113.76,
giving us:
225.6 = 113.76.
The equation is not satisfied; therefore, there seems to be an error in the problem statement. Please check the information provided.
(b) 1 s:
To find the speed and altitude at 1 second, we can use the same equation of motion:
s = ut + (1/2) * a * t².
Substituting the values, we have:
s = 47 * 1 + (1/2) * (-9.8) * (1)².
Simplifying this equation gives us:
s = 47 - 4.9 ≈ 42.1 m.
The speed can be found by using the equation:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Substituting the values, we get:
v = 47 + (-9.8) * 1 = 37.2 m/s.
Therefore, at 1 second, the rocket's speed is 37.2 m/s and its altitude is approximately 42.1 meters.
(c) 4 s:
Following the same steps, we find the altitude by substituting the values into the equation of motion:
s = 47 * 4 + (1/2) * (-9.8) * (4)²,
which simplifies to:
s = 188 - 78.4 ≈ 109.6 m.
For the speed, we use:
v = 47 + (-9.8) * 4 = 7.2 m/s.
At 4 seconds, the rocket's speed is 7.2 m/s, and its altitude is approximately 109.6 meters.
(d) 8 s:
Again, using the equation of motion, we have:
s = 47 * 8 + (1/2) * (-9.8) * (8)²,
which simplifies to:
s = 376 - 313.6 ≈ 62.4 m.
Using the speed equation, we get:
v = 47 + (-9.8) * 8 = -32 m/s.
At 8 seconds, the rocket's speed is -32 m/s (indicating a downward velocity, as expected) and its altitude is approximately 62.4 meters.