A student drops a rock from a bridge to the water 12.0 meters below. With what speed does the rock hit the water?
V^2 = Vo^2 + 2g*h.
Vo = 0.
g = 9.8 m/s^2.
h = 12 m.
V = ?.
1.6s
15.2
Well, let's do some rock-et science! When the student drops the rock, it will accelerate downwards due to gravity. So, using the formulas of constant acceleration, we can find the speed. The only thing I find "deep" here is the water below the bridge!
Now, let's get serious for a moment. To find the speed of the rock, we can use the equation:
v = sqrt(2 * g * d)
Where:
v is the final velocity (speed)
g is the acceleration due to gravity (approximately 9.8 m/s²)
d is the distance (12.0 meters)
Now, let's plug in the values and crunch the numbers.
v = sqrt(2 * 9.8 * 12.0)
v ≈ sqrt(235.2)
v ≈ 15.33 m/s (rounded to two decimal places)
So, the rock will plunge into the water at a speed of approximately 15.33 m/s. Hope that wasn't too dry of an explanation!
To determine the speed at which the rock hits the water, we can use the basic equations of motion. The most appropriate equation for this situation is the equation for the final velocity (vf) of an object in free fall. The equation is:
vf^2 = vi^2 + 2ad
Where:
- vf is the final velocity,
- vi is the initial velocity (which is zero in this case),
- a is the acceleration due to gravity (9.8 m/s^2),
- and d is the distance traveled (12.0 meters).
Since the rock is dropped from rest, the initial velocity (vi) is zero. Plugging in the values we know:
vf^2 = 0 + 2 * 9.8 * 12
Calculating further:
vf^2 = 235.2
To find the final velocity (vf), take the square root of both sides of the equation:
vf ≈ √235.2
vf ≈ 15.34 m/s
Therefore, the rock hits the water with a speed of approximately 15.34 m/s.