Question

A Small Rock Is Thrown Vertically Upward With A Speed Of 29.0 M/S From The Edge Of The Roof Of A 29.0-M-Tall Building. The Rock Doesn't Hit The Building On Its Way Back Down And Lands In The Street Below. Ignore Air Resistance. What Is The Speed Of The Rock Just Before It Hits The Street? Express Your Answer With The Appropriate Units. How Much Time Elapses from when the rock is thrown until it hit a street?

Answer 1

The speed of the rock just before it hits the street is 23.9 m/s.

Step-by-step explanation:

To find the speed of the rock just before it hits the street, we can use the principle of conservation of energy. When the rock is thrown vertically upwards, it gains potential energy. At its highest point, all the initial kinetic energy has been converted into potential energy. As the rock falls back down, it loses potential energy and gains kinetic energy.

Since the potential energy at the highest point is equal to the kinetic energy just before it hits the street, we can equate the two. The potential energy can be expressed as mgh, where m is the mass of the rock, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the building.

Using this equation, we can solve for the velocity of the rock just before it hits the street. From the information given, we know that the height of the building is 29.0 m, so substituting the values:

Potential energy = mgh = m * 9.8 * 29.0 = 284.2m

Since the potential energy is equal to the kinetic energy at the lowest point, we have:

Kinetic energy = 284.2m

Kinetic energy = 1/2 * m * v^2, where v is the velocity of the rock.

Substituting the values:

1/2 * v^2 = 284.2

To solve for v, we can rearrange the equation:

v^2 = 2 * 284.2

v = sqrt(2 * 284.2) = 23.9 m/s

Therefore, the speed of the rock just before it hits the street is 23.9 m/s.