Calculating if a Convertible Gets Wet from Dropped Bucket

[Byrne]

A convertible with the top down is heading for a parking garage entrance at a steady 24 km/h. Directly above this entrance is a window washer who accidentally drops her bucket of water off her rising platform. At this time the car was 12 m from the entrance and the platform was traveling up at a speed of 1 m/s. Calculate if the car gets wet.

Okay, so here is my given information:

CONVERTIBLE:
V(const) = 24 km/h = 6.67 m/s
d = 12 m

WINDOW WASHER
V(initial) = -1.5 m/s

I really don't see where to begin. I know I can determine the time it will take for the convertible to reach the entrance of the garage, but I don't see what good that would do since I do not have the height above the road the window washer is when the bucket falls.

Help.

 

[Tide]

By the way, this is not a kinematics problem. It is a dynamics problem.

Obviously, you need the initial height but if the problem is as you stated then you will take the height, say h, to be a parameter in your analysis.

I think the most direct approach would be to simply determine how long it takes for the bucket to hit the ground (in general this would really be the height of the car). Then calculate where the car would be at that time.

 

[quicknote]

Hello there,

I can provide a response to this scenario. To determine if the convertible gets wet, we need to consider the distance between the convertible and the window washer when the bucket was dropped. As you have mentioned, we do not have the height above the road at which the window washer was when the bucket fell. However, we can make some assumptions and use some equations to calculate this.

Firstly, we can assume that the window washer is at a height of 2 meters above the road when she drops the bucket. This is a reasonable height for a window washer platform to be at. Now, we can use the equations of motion to calculate the time it takes for the bucket to reach the ground. We know that the initial velocity of the bucket is -1.5 m/s and the distance it needs to travel is 2 meters. Using the equation d = Vit + 1/2at^2, we can calculate the time it takes for the bucket to reach the ground, which is approximately 1.6 seconds.

Next, we can use the same equation to calculate the distance between the convertible and the entrance of the garage after 1.6 seconds. We know that the convertible is traveling at a constant speed of 6.67 m/s and the distance it needs to cover is 12 meters. So, using the equation d = Vit + 1/2at^2, we get the distance to be 22.7 meters.

Now, we can calculate the horizontal distance between the convertible and the window washer at the time the bucket reaches the ground. This can be calculated using trigonometry. We know that the distance between the convertible and the entrance of the garage is 22.7 meters and the height at which the window washer dropped the bucket is 2 meters. Using the Pythagorean theorem, we can calculate the horizontal distance to be approximately 22.6 meters.

Finally, we can compare this distance with the horizontal distance between the convertible and the entrance of the garage, which we calculated to be 22.7 meters. Since the horizontal distance between the convertible and the window washer is less than the distance between the convertible and the entrance of the garage, we can conclude that the convertible will not get wet.

In conclusion, by making some reasonable assumptions and using equations of motion and trigonometry, we can determine that the convertible will not get wet from the dropped bucket. However, it is important