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3 years ago

A ball is dropped from a tower that is 206 m high, How long does it take to reach the ground?

Physics

1 answer:

Slav-nsk [51]3 years ago

7 0

It will take 6.42 s for the ball that is dropped from a height of 206 m to reach the ground.

From the question given above, the following data were obtained:

Height (H) = 206 m

NOTE: Acceleration due to gravity (g) = 10 m/s²

The time taken for the ball to get to the ground can be obtained as follow:

H = ½gt²

206 = ½ × 10 × t²

206 = 5 × t²

Divide both side by 5

$t^{2} = \frac{206}{5}\\\\ t^{2} = 41.2$

Take the square root of both side

$t = \sqrt{41.2}$

Therefore, it will take 6.42 s for the ball to get to the ground.

Learn more: brainly.com/question/24903556

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By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

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$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is: