27. Using energy considerations and assuming negligible air resistance,

show that a rock thrown from a bridge 20.0 m above water with an initial
speed of 15.0 m/s strikes the water with a speed of 24.8 m/s independent
of the direction thrown.

(1/2) m Vi^2 + m g h = (1/2) m Vfinal^2

or
Vfinal^2 = Vi^2 + 2 g h

= 225 + 2 (9.81)(20)

= 617.4

sqrt 617.4 = 24.8

I have a fought in the question value of m is not given .

Then how you have found the kinetic energy???

Well, well, well, looks like we've got a gravity-defying rock on our hands here! Let's dive into this problem, shall we?

First things first, we'll start by considering the energy conservation principle. Since we're dealing with negligible air resistance, we can assume that the only forces acting on the rock are gravity and the initial throw. The rock's mechanical energy will be conserved throughout its flight.

Now, as our rock is thrown from the bridge, its initial energy will consist of both kinetic and potential energy. The initial kinetic energy can be calculated using the formula KE = 0.5 * m * v^2, where m is the mass of the rock and v is its initial velocity of 15.0 m/s.

The potential energy at the beginning will be given by PE = m * g * h, where m is the mass of the rock, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the bridge, which is 20.0 m in this case.

Since energy is conserved, the sum of the initial kinetic energy and potential energy should be equal to the final kinetic energy just before hitting the water.

So we have KE_initial + PE_initial = KE_final.

Substituting in the values we have, we get:

0.5 * m * (15.0 m/s)^2 + m * 9.8 m/s^2 * 20.0 m = 0.5 * m * (24.8 m/s)^2

Now, here comes the magical part. If you solve this equation (which I'll leave to you - I'm not a math bot), you'll find that the mass of the rock cancels out completely! It's like a disappearing act, but without the smoke and mirrors.

What does this mean? It means that the final speed of the rock, just before it hits the water, is independent of its mass, the direction it was thrown, or even the type of rock it is.

So, no matter if that rock was thrown by a superhero or a clown (like me!), it will always hit the water with a speed of 24.8 m/s. Isn't that rock-tastic?

To solve this problem, we can utilize the principle of conservation of energy.

The total mechanical energy of the rock is conserved throughout its motion, ignoring any energy losses due to air resistance. The total mechanical energy (E) of an object can be expressed as the sum of its kinetic energy (KE) and potential energy (PE).

E = KE + PE

The kinetic energy of an object is given by the equation:

KE = (1/2) * m * v^2

where m is the mass of the object and v is its velocity.

The potential energy of an object at height h is given by:

PE = m * g * h

where g is the acceleration due to gravity.

For the rock's initial position on the bridge, its potential energy (PE1) is given by:

PE1 = m * g * h1

where h1 is the height of the bridge, and the initial velocity is given as 15.0 m/s.

Next, when the rock strikes the water, its potential energy (PE2) becomes zero, and its kinetic energy (KE2) is given by:

KE2 = (1/2) * m * v^2

where v is the final velocity of the rock.

Since the total mechanical energy is conserved, we can equate the initial and final energies:

PE1 + KE1 = PE2 + KE2

Substituting the expressions for potential and kinetic energies:

m * g * h1 + (1/2) * m * v1^2 = 0 + (1/2) * m * v2^2

Since the mass (m) is common on both sides of the equation, we can cancel it out:

g * h1 + (1/2) * v1^2 = (1/2) * v2^2

Now we can substitute the given values:

g * 20.0 + (1/2) * 15.0^2 = (1/2) * v2^2

Simplify the equation:

9.8 * 20.0 + 0.5 * 225.0 = 0.5 * v2^2

196.0 + 112.5 = 0.5 * v2^2

308.5 = 0.5 * v2^2

Multiply both sides by 2:

617 = v2^2

Take the square root of both sides:

v2 = √617

v2 ≈ 24.8 m/s

Therefore, regardless of the direction thrown, the rock will strike the water with a speed of approximately 24.8 m/s.

24.8