# 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]

##### 1 Answer

#### Answer:

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?*

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

*gravitational acceleration*, equal to

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, **1.2 seconds**, and

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")#