Question

A girl stands on top of the Golden Gate Bridge and drops a penny straight down. How long does it take the penny to fall the 228m to the San Francisco Bay? [FULL QUESTION in the answer]

Answer 1

The penny will hit the water after 6.82 s.

Explanation:

FULL QUESTION

A girl stands on top of the Golden Gate Bridge and drops a penny straight down. How long does it take the penny to fall the 228m to the San Francisco Bay?

What is its velocity of the penny after 1.2 seconds?

How high above the water is it after 1.2 seconds?

The girl's brother stands at the edge of the Golden Gate Bridge and throws a penny straight up at 46.8 km/h. How long does it take the penny to fall to the water, 226m below?

How long does it take the penny to reach its highest point above the water?

What is the penny's maximum height above the water?

What is its velocity after 4.72 seconds have passed?

What is its height after 4.72 seconds have passed?

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So, you know that the girl is standing on top of the Golden Gate Bridge, at a height of 228 m above water.

Since the initial velocity of the penny is euqal to zero, you can say that

`h = 1/2 * g * t^2" "` , where

`h` - the height of the bridge;
`g` - the gravitational acceleration, equal to `"9.8 m s"^(-2)`;
`t` - the time it takes for the penny to hit the water.

Plug in your values to get

`t^2 = (2d)/g implies t = sqrt((2d)/g)`

`t = sqrt((2 * 228color(red)(cancel(color(black)("m"))))/(9.8color(red)(cancel(color(black)("m")))"s"^(-2))) = color(green)("6.82 s")`

To get the penny's velocity after 1.2 seconds, you need to use the equation

`v = v_0 + at`

In your case, `a` is equal to `g`, `t` will be equal to 1.2 seconds, and `v` will be the velocity of the penny after 1.2 seconds.

Plug in your values to get

`v_1.2 = g * t_1.2`

`v_1.2 = 9.8 "m"/"s"^color(red)(cancel(color(black)(2))) * 1.2color(red)(cancel(color(black)("s"))) = color(green)("11.76 m/s")`

To get the penny's heigh above the water after 1.2 seconds, calculate the distance it covered in that much time, then subtract this distance from the height of the bridge.

`h_1.2 = 1/2 * g * "1.2 s"^2 = 1/2 * 9.8"m"/color(red)(cancel(color(black)("s"^2))) * 1.2^2 color(red)(cancel(color(black)("s"^2))) = "7.06 m"`

This means that the penny's heigh above the water will be

`h_"above water" = 228 - 7.06 = color(green)("221 m")`

Now, before you calculate how long it takes for the penny her brother threw to reach the water, you need to convert its initial velocity from km/h to m/s.

`46.8color(red)(cancel(color(black)("km")))/color(red)(cancel(color(black)("h"))) * (1color(red)(cancel(color(black)("h"))))/("3600 s") * ("1000 m")/(1color(red)(cancel(color(black)("km")))) = "13 m/s"`

Now, determine the height the penny travels upwards

`v^2 = v_0^2 - 2 * g * h_"up"`

At maximum height, the penny's velocity will be equal to zero. This means that you have

`0 = v_0^2 - 2g * h_"up" implies h_"up" = v_0^2/(2g) = (13^2"m"^color(red)(cancel(color(black)(2))) color(red)(cancel(color(black)("s"^(-2)))))/(2 * 9.8color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-2))))) = "8.62 m"`

The maximum height of the penny will be

`h_"max" = 228 + 8.62 = color(green)("236.6 m")`

The time it takes the penny to reach this height is

`v = v_0 - g * t_"up"`

`t_"up" = v_o/g = (13color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(9.8 color(red)(cancel(color(black)("m"))) "s"^color(red)(cancel(color(black)(-2)))) = color(green)("1.33 s")`

Now calculate the time it takes the penny to fall from its maximum height

`h_"max" = 1/2 * g * t_"down"^2 implies t_"down" = sqrt((2 h_"max")/g)`

`t_"down" = sqrt((2 * 236.6color(red)(cancel(color(black)("m"))))/(9.8color(red)(cancel(color(black)("m"))) "s"^(-2))) = "6.95 s"`

The total time it takes the penny to reach the water, from the moment it was thrown by the girl's brother, will be

`t_"total" = t_"up" + t_"down"`

`t_"total" = 1.33 + 6.95 = color(green)("8.28 s")`

To calculate its velocity after 4.72 seconds have passed, use the fact that it takes 1.33 seconds for it to go up. This means that after 1.33 seconds it will be falling. The time will be

`t_"fall" = 4.72 - 1.33 = "3.39 s"`

Now simply use the equation

`v_"fall" = g * t_"fall" = 9.8"m"/"s"^color(red)(cancel(color(black)(2))) * 3.39color(red)(cancel(color(black)("s"))) = color(green)("33.2 m/s")`

and

`h_"fall" = 1/2 * g * t_"fall"^2`

`h_"fall" = 1/2 * 9.8"m"/color(red)(cancel(color(black)("s"^2))) * 3.39^2 color(red)(cancel(color(black)("s"^2))) = "56.3 m"`

This means that its height above the water after 4.72 seconds will be

`h_"above" = 236.6 - 56.3 = color(green)("180.3 m")`