A rocket moves straight upward, starting from rest with an acceleration of +25.5 m/s2. It runs out of fuel at the end of 4.26 s and continues to coast upward, reaching a maximum height before falling back to Earth.
(a) Find the rocket's velocity and position at the end of 4.26 s.
vb = m/s
yb = m
(b) Find the maximum height the rocket reaches.
m
(c) Find the velocity the instant before the rocket crashes on the ground.
m/s
a. Vf=at=25.5 m/s^2 * 4.26s=10=108.6m/s
h=0.5a*t^2 = 0.5*25.5*(4.26)^2=231.4 m
b. hmax = (Vf^2-Vo^2) / 2g,
hmax = (0-(108.6)^2)/ -19.6 = 601.7 m.
c. Vf^2 = Vo^2 + 2g*d ,
Vf^3 = 0 + 19.6*601.7 = 11793.3,
Vf = 108.6.
To solve this problem, we can use the equations of motion to find the rocket's velocity and position at the given time, the maximum height it reaches, and the velocity right before it crashes onto the ground.
Let's break down each part of the problem one by one:
(a) Finding the rocket's velocity and position at the end of 4.26 s:
1. The initial velocity of the rocket, u, is 0 since it starts from rest.
2. The acceleration, a, is +25.5 m/s^2, since it's moving straight upward.
3. The time, t, is 4.26 s.
We can use the equation of motion: v = u + at
Substituting the given values:
v = 0 + (25.5 m/s^2)(4.26 s)
v = 108.63 m/s
Therefore, the rocket's velocity at the end of 4.26 s is 108.63 m/s.
Now, let's find the position using the equation: s = ut + (1/2)at^2
Since the rocket starts from rest, the initial position, y0, is also 0.
Substituting the given values:
s = 0 + (1/2)(25.5 m/s^2)(4.26 s)^2
s = 230.78 m
Therefore, the rocket's position at the end of 4.26 s is 230.78 m above the ground.
(b) Finding the maximum height the rocket reaches:
To find the maximum height, we need to find the time it takes for the rocket to reach its highest point. After this point, it will start falling back down.
Since the rocket's velocity will become 0 at the highest point, we can use the equation: v = u + at
Substituting the given values:
0 = 0 + (25.5 m/s^2)t_max
t_max = 0 s (since the velocity is 0 at the highest point)
The rocket takes 4.26 s to reach its highest point. Therefore, the maximum height it reaches is the same as the position at the end of 4.26 s, which is 230.78 m.
(c) Finding the velocity right before the rocket crashes onto the ground:
Since we know the rocket reaches a maximum height and then falls back to the ground, the time it takes to reach the ground is double the time it took to reach the maximum height (4.26 s * 2 = 8.52 s).
The acceleration due to gravity, g, is approximately -9.8 m/s^2 (assuming the rocket is in a vacuum).
The initial velocity, u, is the velocity at the highest point, which is 0.
We can use the equation of motion: v = u + at
Substituting the given values:
v = 0 + (-9.8 m/s^2)(8.52 s)
v = -83.496 m/s (rounding to three significant figures)
Therefore, the velocity right before the rocket crashes onto the ground is approximately -83.5 m/s (negative sign indicates downward direction).
Please note that this solution assumes no air resistance and a constant acceleration due to gravity.