A spy in a speed boat is being chased down a
river by government officials in a faster craft.
Just as the officials’ boat pulls up next to the
spy’s boat, both boats reach the edge of a 4.6
m waterfall. The spy’s speed is 16 m/s and
the officials’ speed is 26 m/s.
How far apart will the two vessels be when
they land below the waterfall? The acceleration
of gravity is 9.81 m/s
2
.
To determine how far apart the two vessels will be when they land below the waterfall, we need to calculate the time it takes for each boat to fall from the top of the waterfall to the bottom.
First, let's find the time it takes for the spy's boat to fall.
Using the equation of motion:
h = (1/2) * g * t^2
Where:
h = height of the waterfall = 4.6 m
g = acceleration due to gravity = 9.81 m/s^2
t = time
Rearranging the equation, we get:
t^2 = (2h) / g
t^2 = (2 * 4.6) / 9.81
t^2 = 0.9392
t = √0.9392
t ≈ 0.969 s
Now, let's find the time it takes for the officials' boat to fall. Since they are already at the edge of the waterfall, they will fall for the same amount of time as the spy's boat.
Therefore, both boats will fall for approximately 0.969 seconds.
Next, let's calculate the distance each boat travels horizontally during that time.
For the spy's boat:
distance = speed * time
distance = 16 m/s * 0.969 s
distance ≈ 15.504 m
For the officials' boat:
distance = speed * time
distance = 26 m/s * 0.969 s
distance ≈ 25.194 m
Therefore, when they land below the waterfall, the two boats will be approximately 15.504 m + 25.194 m = 40.698 m apart.
To find the distance between the two vessels when they land below the waterfall, we need to first calculate the time it takes for each vessel to fall off the waterfall.
We can use the equation of motion for an object in free fall to find the time it takes for an object to fall a certain distance. In this case, the distance is 4.6 m and the acceleration due to gravity is 9.81 m/s^2.
For the spy's boat:
Using the equation h = (1/2)gt^2, where h is the distance (4.6 m), g is the acceleration due to gravity (9.81 m/s^2), and t is the time, we can solve for t.
Rearranging the equation, we get t = sqrt(2h/g).
Plugging in the values, we get t = sqrt((2 * 4.6) / 9.81) seconds.
For the officials' boat:
Using the same equation, h = (1/2)gt^2, with h being the same distance (4.6 m), g is still the acceleration due to gravity (9.81 m/s^2), and t is the time.
Plugging in the values, we get t = sqrt((2 * 4.6) / 9.81) seconds.
Since both boats are falling off the waterfall at the same time, the time will be the same for both vessels.
Now, we can calculate the distance between the two vessels when they land below the waterfall using the equation d = v * t, where d is the distance, v is the relative velocity between the two vessels, and t is the time it takes for them to fall off the waterfall.
The relative velocity between the two vessels can be calculated by subtracting the speed of the spy's boat from the speed of the officials' boat. So v = 26 m/s - 16 m/s.
Plugging in the values, we get d = (26 - 16) * t.
Now, we have all the necessary information to calculate the distance between the two vessels:
1. Calculate t = sqrt((2 * 4.6) / 9.81)
2. Calculate d = (26 - 16) * t
By following these steps, you can find the distance between the two vessels when they land below the waterfall.
Same answer.
h = 4.6 m.
V1 = 16 m/s.
V2 = 26 m/s.
g = 9.81 m/s^2
h = 0.5g*t^2
Solve for t = Fall time.
d = (V2-V1)*t = Distance apart.