The engine of a toy rocket supplies an average acceleration of 38.0 m/s/s to the rocket for an interval of 0.80 s. A.)If the toy roocket is launched vertically, how high does it rise in this interval? B.) How fast is the rocket moving at the end of 0.80 s? C.) What altitude does the rocket reach before falling back to Earth? D.) How long does it take the rocket to reach this altitude?
A) Solve Y = (a/2) T^2.
a is the acceleration and T is the 0.80 s rocket burn time.
B) Solve
V = a T, using the T = 0.8 s
V is the rocket velocity at 0.8 s.
C) Add Y (from part (A) ) and the additional distance that it rises after burnout, Y'.
After burnout, it continues to rise until (1/2) V^2 = g Y'.
Solve for the additiona altitude gain, Y'. The total altitude at that time is Y' + Y
D) 0.8 s PLUS the time it rises after burnout, T'.
g T' = V
T' = V/g
Total time rising = T + T'.
16.12
A.) To determine the height the rocket rises in this interval, we can use the equation for vertical motion:
h = (1/2) * a * t^2
where:
h = height
a = acceleration
t = time
In this case, the acceleration (a) is given as 38.0 m/s^2, and the time (t) is given as 0.80 s. Plugging in these values into the equation, we get:
h = (1/2) * 38.0 m/s^2 * (0.80 s)^2
h = (1/2) * 38.0 m/s^2 * 0.64 s^2
h = 12.16 m
Therefore, the rocket rises to a height of 12.16 meters in this interval.
B.) To determine the rocket's speed at the end of 0.80 s, we can use the equation for velocity:
v = a * t
where:
v = velocity
a = acceleration
t = time
In this case, the acceleration (a) is still 38.0 m/s^2, and the time (t) is 0.80 s. Plugging in these values into the equation, we get:
v = 38.0 m/s^2 * 0.80 s
v = 30.4 m/s
Therefore, the rocket is moving at a speed of 30.4 m/s at the end of 0.80 s.
C.) To determine the altitude the rocket reaches before falling back to Earth, we need to consider that the rocket will reach its highest point when its velocity becomes zero. At this point, the rocket's potential energy will be maximum. We can use the equation:
v^2 = u^2 + 2a(h - h0)
where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (30.4 m/s)
a = acceleration (-9.8 m/s^2, considering downward as negative)
h = final altitude (unknown)
h0 = initial altitude (0 m, assuming the rocket starts from the ground)
Substituting these values into the equation, we get:
0^2 = (30.4 m/s)^2 + 2 * (-9.8 m/s^2) * (h - 0)
0 = 924.16 + (-19.6h)
19.6h = 924.16
h = 924.16 / 19.6
h = 47.14 m
Therefore, the rocket reaches an altitude of 47.14 meters before falling back to Earth.
D.) To determine the time it takes for the rocket to reach this altitude, we can use the equation for vertical motion:
h = u * t + (1/2) * a * t^2
Since we want to find the time (t), we can rearrange the equation:
(1/2) * a * t^2 + u * t - h = 0
Plugging in the values, we get:
(1/2) * (-9.8 m/s^2) * t^2 + 30.4 m/s * t - 47.14 m = 0
Solving this quadratic equation will give us the time it takes for the rocket to reach the altitude of 47.14 m. The solutions will be the time at the ascending phase and the time at the descending phase. However, for simplicity, we will consider the positive solution:
t = 3.04 s
Therefore, it takes 3.04 seconds for the rocket to reach the altitude of 47.14 meters.
To solve these problems, we can use the equations of motion for constant acceleration. There are four key equations that can help us solve these types of problems:
1. v = u + at
2. s = ut + (1/2)at^2
3. v^2 = u^2 + 2as
4. s = vt - (1/2)at^2
Where:
- v is the final velocity of the rocket
- u is the initial velocity of the rocket
- a is the acceleration of the rocket
- t is the time interval
- s is the displacement (change in position) of the rocket
Let's solve each part of the problem step by step:
A) To find the height the rocket rises in this interval, we need to find the displacement of the rocket (s). We are given the average acceleration (a) and the time interval (t).
Using equation 2, s = ut + (1/2)at^2, we can substitute the given values:
s = 0 + (1/2)(38.0 m/s^2)(0.80 s)^2
s = 0 + 15.2 m/s^2 * 0.64 s^2
s = 9.728 m
Therefore, the rocket rises approximately 9.728 meters in this interval.
B) To find the velocity of the rocket at the end of 0.80 s, we need to use equation 1, v = u + at.
Using the same given values as before, we can substitute:
v = 0 + (38.0 m/s^2)(0.80 s)
v = 30.4 m/s
Therefore, the rocket is moving at a speed of 30.4 m/s at the end of 0.80 s.
C) To find the altitude the rocket reaches before falling back to Earth, we need to find the maximum height achieved. At this point, the velocity will be zero, and by symmetry, the time it takes to reach this point is half of the total time.
Using equation 1, v = u + at, we can set v = 0 and solve for t:
0 = 38.0 m/s^2 * t
t = 0
Since the velocity is zero at time t = 0, the rocket reaches its maximum height and starts falling back to Earth.
D) The time it takes for the rocket to reach this maximum altitude is half of the total time. We are given the total time interval as 0.80 s, so the time to reach the maximum altitude would be half of that, i.e., 0.80 s / 2 = 0.40 s.
Therefore, it takes the rocket 0.40 seconds to reach its maximum altitude.