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Crazy boy [7]
4 years ago
10

An object is dropped from 26 feet below the tip of the pinnacle atop a 702 ft tall building. The height h of the object after t

seconds is given by the equation h=-16t^2+676. Find how many seconds pass before the object reaches the ground.
Physics
1 answer:
Dimas [21]4 years ago
4 0

Answer:

6.5 seconds.

Explanation:

Given: h=-16t²+679

When the object reaches the ground, h=0.

∴ 0=-16t²+679

collecting like terms,

⇒ 16t²=679

Dividing both side of the equation by the coefficient of t² i.e 16

⇒ 16t²/16 = 679/16

⇒ t² = 42.25

taking the square root of both side of the equation.

⇒ √t² =√42.25

⇒ t = 6.5 seconds.

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On Earth, the rocket's weight is  (0.8kg) x (9.8 m/s²) = 7.84 newtons

On the Moon, the rocket's weight is  (0.8kg) x (1.62 m/s²) = 1.3 newtons

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Now we have all the data we need to mash together and calculate the
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                           Net Force = (mass) x (acceleration).
 
Since the question is asking for acceleration, let's first solve Newton's law
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On Earth, the forces on the rocket are

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On the moon, the forces on the rocket are

         (weight of 1.3 N down) + (blast of 12 N up) = 10.7 newtons UP (net)

         Acceleration = (10.7 newtons UP) / (0.8 kg) = 13.375 m/s² UP   
          
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