EugenenebriI000idn
Answered

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The distance travelled by a falling object varies directly as the square of the time it spends in falling. If a body falls 64ft in 2 seconds, find the distance that it will fall in 10 seconds.​

Sagot :

Martusjenylynidn

Answer:

The distance travelled by a falling object is proportional to the square of the time it spends falling, according to the equation d = 1/2 * g * t^2, where d is the distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.

Given:

  • The object falls 64 ft in 2 seconds.
  • We need to find the distance it will fall in 10 seconds.

Using the equation d = 1/2 * g * t^2:

  • At 2 seconds, the distance fallen is 64 ft.
  • Plugging in the values, we get: 64 = 1/2 * 9.8 * 2^2
  • Solving for g, we get g = 9.8 m/s^2

Now, to find the distance fallen in 10 seconds:

  • d = 1/2 * 9.8 * 10^2 = 490 ft

Therefore, the distance the object will fall in 10 seconds is 490 ft.

NoahBadillaidn

DISTANCE

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[tex]\rm{\underline{PROBLEM:}}[/tex]

The distance travelled by a falling object varies directly as the square of the time it spends in falling. If a body falls 64ft in 2 seconds, find the distance that it will fall in 10 seconds.​

[tex]\\[/tex]

[tex]\rm{\underline{SOLUTION:}}[/tex]

The formula for the distance traveled by a falling object while varying directly with the square of the time the object falls is:

[tex]\large{\boxed{d = kt^{2} }}[/tex]

  • d - is the distance
  • t - is the time
  • k - is the constant of variation

[tex]\\[/tex]

Substitute the values into the formula to solve for the value of k :

  • [tex]{d = kt^{2} }[/tex]
  • [tex]{64 = k(2)^{2} }[/tex]
  • [tex]{64 = k \times 4}[/tex]
  • [tex]k = \frac{64}{4}[/tex]
  • [tex]k = 16[/tex]

[tex]\\[/tex]

Now that we have the value of the constant k, find the distance traveled of the falling object in 10 seconds. Substitute k = 16 and t = 10.

  • [tex]{d = kt^{2} }[/tex]
  • [tex]{d = 16(10)^{2} }[/tex]
  • [tex]{d = 16 \times 100}[/tex]
  • [tex]d = \underline{\green{1,600}}[/tex]

Hence, the distance traveled of the falling object is 1,600 feet.

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[tex]\large{\boxed{\tt{\blue{06/24/2024}}}}[/tex]

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