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kramer
3 years ago
11

A small rocket with mass 20.0 kg is moving in free fall toward the earth. Air resistance can be neglected. When the rocket is 80

.0 m above the surface of the earth, it is moving downward with a speed of 30.0 m/s. At that instant the rocket engines start to fire and produce a constant upward force F on the rocket. Assume the change in the rocket's mass is negligible. What is the value of F if the rocket's speed becomes zero just as it reaches the surface of the earth, for a soft landing? (Hint: The net force on the rocket is the combination of the upward force F from the engines and the downward weight of the rocket.)
Physics
1 answer:
Mekhanik [1.2K]3 years ago
7 0

Answer:

= 308.5 N

Explanation:

acceleration  of rocket for safe landing

v_{y} ^{2} = v_{0} ^{2} +2ay

a = \frac{v_{y}^{2} - v_{0}^{2}  }{2y}

v_{0} = initial velocity

v_{y} =final Velocity

m = mass of rocket

\frac{0^{2} - 30^{2}  }{2\times80}

-5.625 m/s^{2}

Upward force

F - mg = ma

F = ma+ mg

F = m(a+g)

m= mass

a = 5.625 m/s^{2}

g = 9.8m/s^{2}

= 20(5.625 m/s^{2}+ 9.8m/s^{2})

= 308.5 N

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3 years ago
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<h2>Answer: 10.52m</h2><h2 />

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V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}   (4)

As this is a projectile motion, we have two principal equations related:

<h2>In the x-axis: </h2>

X=V_{ox}.t  (5)

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If we already know X and V_{ox}, we have to find the time (we will need it for the following equation):

t= \frac{X}{ V_{ox}}  (6)

t=2.42s  (7)

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-y=V_{oy}.t+\frac{1}{2}g.t^{2}   (8)

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y is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

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Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

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Multiplying by -1 each side of the equation:

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Answer:

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anothe way to do it is,

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