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padilas [110]
3 years ago
8

A model rocket is launched straight upward with an initial speed of 56.5 m/s. It accelerates with a constant upward acceleration

of 1.96 m/s 2 until its engines stop at an altitude of 198.8 m. What is the maximum height reached by the rocket? Answer in units of m.
Physics
1 answer:
marta [7]3 years ago
3 0

Answer:

Maximum height reached by the rocket, h = 202.62 meters            

Explanation:

It is given that,

Initial speed of the model rocket, u = 56.5 m/s

Constant upward acceleration, a=1.96\ m/s^2

Distance traveled by the engine until it stops, d = 198.8 m

Let v is the speed of the rocket when the engine stops. It can be calculated using the third equation of motion as :

v=\sqrt{u^2+2ad}

v=\sqrt{(56.5)^2+2\times 1.96\times 198.8}    

v = 63.02 m/s

At the maximum height, v = 0 and the engine now decelerate under the action of gravity, a = -g. Let h is the maximum height reached by the rocket.

Again using third equation of motion as :

v^2-u^2=-2gh

h=\dfrac{v^2-u^2}{-2g}

h=\dfrac{u^2}{2g}

h=\dfrac{(63.02)^2}{2\times 9.8}

h = 202.62 meters

So, the maximum height reached by the rocket is 202.62 meters. Hence, this is the required solution.

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Answer:

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Explanation:

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Using the eqn of motion

{v}^{2}  -  {u}^{2}  = 2gh

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So,

{30}^{2}  -  {0}^{2}  = 2 \times 10 \times h

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1:04-1:10 hours

Explanation:

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  3. Now you have to go to the second part of your table and look for the distance column, in feet, of your second dive. We find 60 and then going right in the blue row, we'll look for the time (35) or its closest value (36).
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1:-

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\\ \sf\longmapsto 45.23L

\\ \sf\longmapsto 45.23\times 1000

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2:-

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\\ \sf\longmapsto .035hL

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3:-

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\\ \sf\longmapsto 27.32

\\ \sf\longmapsto 27.32\times 0.0001

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