Question

A 2.00 kg ball is thrown straight upward from the top of a 50.0 m high building with an initial speed of 10.0 m/s.

       Given:

       a. What is the total energy at the top of the building?

       b. What is the total energy at the ground?

       c. What are the potential and kinetic energies at the ground?

       d. What is its speed at the ground?

Answer 1

Final Answer:-

a) 1081 Joules

b) 1081 Joules

c) potential energy at ground = 0 J

Kinetic energy at ground = 1081 J

d) 32.88 m/s

Solution:-

As total mechanical energy is given by,

E = Kinetic Energy + Potential Energy

E = (0.5 * m * v²) + ( m* g *h)

at the top of the building,

Et = (0.5 * 2 * 10²) + (2 * 9.81 * 50)

Et = 100 + 981 = 1081 Joules

at the ground,

potential energy at ground,

PEg = 2 * 9.81 * 0 [height from the ground is zero]

PEg = 0 Joules

Kinetic energy at ground,

first calculate the distance ball will go upward from the building,

Using third equation of motion,

v² - u² = 2*g*s

0 - 10² = 2 * (-9.81)*s

s = 5.10 meters

Now the ball will travel 50 + 5.1 = 55.1 meters

so,

Vg² - 0 = 2 * 9.81 * 55.1

Vg = 32.88 m/s

Thus kinetic energy at ground,

KE = 0.5 * 2 * 32.88²

KE = 1081 Joules

Total energy of the ball in ground,

E = 0 + 1081 = 1081 Joules

* Which also shows the conservation of energy.