Will the driver get wet?

Question

1.   A convertible with its top down drives towards the entrance of an underground garage at a velocity of 24 km/h. A window cleaner directly above the entrance drops a bucket of water. If the window cleaner is 9.0 m above the entrance and moving upwards at 1.5 m/s and at the moment he drops the bucket the car is 12 m from the entrance, will the driver get wet?

Answer 1

Hi Fawn:

The method here is to determine whether the falling bucket and the moving car are at the same place at the same time, in which case the driver gets wet, or not.

We aren't given the height of the driver in the car above the ground, so let's assume the cleaner's height is taken from the top of the driver's head.

First, let's apply the laws of motion to the bucket. Initially it the bucket moves upward with the speed of the cleaner, until it reaches a speed of zero at the top and falls to the ground, 9 m below.

s = v₀t + (1/2)at² where s = distance traveled, v₀ = initial speed, a = acceleration, and t = time

9 = -1.5t + 0.5 x 9.8t² . . . . . . note that v₀ is negative, since it's in the opposite direction to gravity 4.9t² - 1.5t - 9 = 0 . . . . this is a quadratic in t, so solve for t using the quadratic formula

t = (1.5 +-√(1.5² + 4 x 4.9 x 9)) / (2 x 4.9)

= 1.5 seconds . . . . . . . ignoring the negative answer

Now, we find where the car is after the same time. The car's acceleration isn't given so we assume it's zero.

So, for the car: s = v₀t + (1/2)at²

= 24 x 1.5 + 0

= 36 meters

So, the car is well past the entrance and the driver stays toasty and dry.