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
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To find the rocket's velocity and position at the end of 4.26 seconds, we can use the equations of motion for constant acceleration.
(a) Velocity at the end of 4.26 seconds (vb):
We know that the initial velocity (v0) is 0 m/s (starting from rest), the acceleration (a) is +25.5 m/s^2, and the time (t) is 4.26 seconds.
The equation to find the final velocity (v) is:
v = v0 + at
Substituting the given values:
vb = 0 + (25.5 m/s^2) * 4.26 s
vb = 108.63 m/s
Therefore, the rocket's velocity at the end of 4.26 seconds is 108.63 m/s.
Position at the end of 4.26 seconds (yb):
To find the position, we need to use the equation:
y = y0 + v0t + (1/2)at^2
Since the rocket starts from rest, the initial position (y0) is 0 m, and its initial velocity (v0) is also 0 m/s. Plugging in the given values, we get:
yb = 0 + 0 + (1/2)(25.5 m/s^2)(4.26 s)^2
yb = 227.23 m
Therefore, the rocket's position at the end of 4.26 seconds is 227.23 m.
(b) Maximum height reached by the rocket:
To find the maximum height, we need to determine the time at which the rocket reaches its peak. This occurs when the rocket's velocity becomes 0 m/s.
Using the equation:
vb = v0 + at
0 = 108.63 m/s + (-25.5 m/s^2)t
Solving for t, we find:
t = 4.26 seconds
Now we can find the maximum height (ymax) using the equation:
ymax = y0 + v0t + (1/2)at^2
Since the rocket's initial position (y0) is 0 m and its initial velocity (v0) is 0 m/s, we have:
ymax = 0 + 0 + (1/2)(25.5 m/s^2)(4.26 s)^2
ymax = 275.15 m
Therefore, the maximum height reached by the rocket is 275.15 m.
(c) Velocity at the instant before crashing on the ground:
To find the velocity at the instant before the rocket crashes, we will use the equation:
v^2 = v0^2 + 2a(y - y0)
Since the rocket is now falling downward, the initial position (y0) is the maximum height reached (yb = ymax = 275.15 m), its initial velocity (v0) is 0 m/s, and the acceleration (a) is -9.8 m/s^2 (due to gravity). Plugging in the values, we have:
v^2 = 0^2 + 2(-9.8 m/s^2)(275.15 m - 0)
v^2 = 2(-9.8 m/s^2)(275.15 m)
Calculating the value, we find:
v^2 = -5394.77 m^2/s^2
Since we know the velocity value will be negative (as the object is moving downward), the velocity is:
v = -73.47 m/s
Therefore, the velocity of the rocket at the instant before crashing on the ground is -73.47 m/s.