A key falls from a bridge that is 45 m above the water. The key falls straight down and lands in a model boat traveling at a constant velocity of 3.5 m/s. Calculate the distance the boat was from the point of impact when the key was released.

Answer d= 10.6 m

distance=boatspeed*timeoffall.

To solve this problem, we need to calculate the time it takes for the key to fall from the bridge to the water, and then use the boat's velocity to determine the distance it traveled during that time.

First, let's calculate the time it takes for the key to fall from the bridge to the water. We can use the equation for free fall:

h = (1/2) * g * t^2

where h is the height of the bridge (45 m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time it takes.

Rearranging the equation to solve for t:
t = sqrt((2 * h) / g)

Substituting the values:
t = sqrt((2 * 45) / 9.8) = sqrt(90 / 9.8) = sqrt(9.1837) ≈ 3.03 s

Now that we have the time it takes for the key to fall, we can calculate the distance the boat traveled during that time. We can use the formula:

d = v * t

where d is the distance traveled by the boat, v is the velocity of the boat (3.5 m/s), and t is the time calculated earlier (3.03 seconds).

Substituting the values:
d = 3.5 * 3.03 ≈ 10.6 m

Therefore, the distance the boat was from the point of impact when the key was released is approximately 10.6 meters.