A student drops a rock from a bridge to the water 21.0 m below. With what speed does the rock strike the water

Answer 1

This problem can be solved a number of ways.
One way, would be to employ the law of conservation of energy.
The rock initially has some amount of gravitational Potential Energy (PE) when it is stilling on the bridge. As the rock falls its PE in converted into Kinetic Energy (KE). When the rock hits the ground, all of its PE has been converted into KE.

PE = mgh
KE = 1/2 mv^2
Where m is the rock's mass, g is the gravitational acceleration, h is the height of the bridge, and v is the final velocity of the rock as it hits the ground. You might notice that we are not given mass in problem; however, this is alright since when we set these two equations equal to each other,
mgh = 1/2 mv^2
mass cancels out,
gh = 1/2 v^2
Now all we do is solve for v,
v = sqrt (2*g*h)

You are given a value for h (21.0 meters), and we can assume that the value of g is 9.81 m/s^2.
Simply plug in and solve.
The answer I calculate is 20.3 m/s.

Another possible solution to this problem involves using the kinematics equations to solve for the final velocity of the rock after experiences a constant acceleration over a distance,
v_f^2 = v_i^2 + 2ad
Where v is the velocity (f and i stand for final and initial respectively), a is the constant acceleration the rock receives, and d is the distance over which the rock accelerates.
In this case, the initial velocity is zero and a = g = 9.81 m/s^2.

You could also do it another way,
d = 1/2 at^2
Solve for t, and then plug into,
v = at
Where a is the constant, gravitational acceleration, d is the distance the rock travels, t is the time, and v is the final velocity.

All of these procedures will give you exactly the same solution in the end.

Answer 2

Use right here equations: vf^2 = vi^2 + 2*a*d d = vi*t + a million/2*a*t^2 for the 1st, the preliminary velocity is 0, so vf^2 = 0 + 2*9.8*10.3 v^2 = 201.88 so v = 14.21 m/s for the 2d, preliminary velocity is lower back 0, so 4.2 = 0 + .5*a million.sixty 4*t^2 t^2 = 5.122 and t = 2.26 seconds

Answer 3

That's a good question, but what is the weight of the rock?