A student drops a rock from a bridge to the water 10.1 m below. The acceleration of gravity is 9.8 m/s^2. With what speed does the rock strike the water?

Answer in units of m/s.

V^2 = Vo^2 + 2g*d = 0 + 19.6*10.1 = 198

V = 14.1 m/s.

To find the speed at which the rock strikes the water, we can use the equation of motion:

v^2 = u^2 + 2as

Where:
v = final velocity (speed at which the rock strikes the water)
u = initial velocity (which is 0 in this case since the rock is dropped)
a = acceleration (which is the acceleration due to gravity, -9.8 m/s^2 as it acts downwards)
s = displacement (the distance the rock falls, which is 10.1 m)

Plugging in the values:

v^2 = 0^2 + 2*(-9.8) * 10.1

v^2 = 0 + (-196) * 10.1

v^2 = -1987.6

Taking the square root of both sides to solve for v:

v = √(-1987.6)

Since the velocity must be positive, the negative sign can be ignored.

v ≈ 44.6 m/s

Therefore, the speed at which the rock strikes the water is approximately 44.6 m/s.

To find the speed at which the rock strikes the water, we can use the kinematic equation:

v^2 = u^2 + 2as

where:
v is the final velocity (the speed at which the rock strikes the water)
u is the initial velocity (which is zero for a freely falling object)
a is the acceleration (the acceleration due to gravity)
s is the displacement (the distance the rock falls, which is 10.1 m)

Since the initial velocity is zero, we can simplify the equation to:

v^2 = 2as

Plugging in the values, we have:

v^2 = 2 * 9.8 m/s^2 * 10.1 m

v^2 = 196.98 m^2/s^2

To find v, we can take the square root of both sides of the equation:

v = sqrt(196.98 m^2/s^2)

v ≈ 14.04 m/s

Therefore, the rock strikes the water with a speed of approximately 14.04 m/s.

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