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Pani-rosa [81]
3 years ago
11

A ball is thrown into the air with an upward velocity of 20 feet per second. Its height, h, in feet after t seconds is given by

the function h(t)=-16t^2+20t+2. What is the ball's maximus height? How long does it take the ball to reach its maximum height? Round to the nearest hundredth, if necessary.
A. Reaches a max height of 8.25 feet after 1.25 seconds
B. Reaches a max height of 8.25 feet after 0.63 seconds
C. Reaches a max height of 0.16 feet after 1.34 seconds
D. Reaches a max height of 0.32 feet after 1.34 seconds.
I will give medal thanks!
Physics
1 answer:
Margarita [4]3 years ago
3 0
Thank you for posting your question here at brainly. Feel free to ask more questions.  
<span>The best and most correct answer among the choices provided by the question is B. Reaches a max height of 8.25 feet after 0.63 seconds</span> .     <span><span>

</span><span>Hope my answer would be a great help for you. </span> </span>  

<span> </span>

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n astronaut who weighs 800 N on the surface of the earth lifts off from planet Zuton in a space ship. The free-fall acceleration
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Answer: 0.29 kN

Explanation:

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g_{E}=9.8 m/s^{2} is the free fall acceleration due gravity on Earth (directed downwards)

g_{Z}=3 m/s^{2} is the free fall acceleration due gravity on Zuton (directed downwards)

a=0.5 m/s^{2} is the acceleration of the spaceship at litoff (directed upwards)

We have to find the <u>magnitude of the force</u> F the space ship exerts on the astronaut.

Firstly, we have to know weight has a direct relation with the mass and the acceleration due gravity. In the case of Earth is:

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Where m is the mass of the atronaut.

Isolating m:

m=\frac{W_{E}}{g_{E}} (2)

m=\frac{800 N}{9.8 m/s^{2}} (3)

m=81.63 kg (4)

Now that we know the mass of the astronaut, we can find its weight on Zuton:

W_{Z}=mg_{Z} (5)

W_{Z}=(81.63 kg)(3 m/s^{2}) (6)

W_{Z}=244.89 N (7)

Then, we can calculate the force the space ship exerts on the astronaut by the following equation:

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Isolating F:

F=m.a+W_{Z} (9)

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F=285.7 N \frac{1 kN}{1000 N}=0.285 kN (11)

Finally:

F=0.285 kN \approx 0.29 kN

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