Question

A 6 kg bowling ball is dropped from the top of a building. If it hits the ground with a speed of 55.0 m/s, how much PE did it have at the top of the building and how tall was the building?​

Answer 1

✰ Question Given :

• ➦ A 6 kg bowling ball is dropped from the top of a building. If it hits the ground with a speed of 55.0 m/s, how much PE did it have at the top of the building and how tall was the building?

✰ Required Solution :

★ Value Given to us :

• ⟿ Mass of ball = 6kg

• ⟿ Speed of hitting to ground = 55 m/s

★ Formula used Here :

• ⟿ P.E = mgh

• ⟿ K.E = 1 / 2 mv²

★ According to Question :

• ⟿ P.E = mgh

• ⟿ P.E = 6 × 10 × h

• ⟿ P.E = 60h _____ (eq1)

★ Finding Kinectic Energy :

• ⟿ K.E = 1/2 mv²

• ⟿ K.E = 1/2 × 6 × (55) ²

• ⟿ K.E = 9075J _____(eq2)

★ Evaluation of Eq 1 , 2 :

• ➦ K.E = P.E

• ➦ 9075 J = 60h

• ➦ h = 151.25 m

✰ Height of Building is 151.25 m ✰

_____________________________

Answer 2

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Well, our toy has an energy E, which is the sum of the potential and kinetic energy.So, when the toy is at the top, 7 m, it has potential energy, because it is at rest (its speed is 0)

Ep = m*g*h

Where m is the mass (8 kg), g is the Earth's gravity (9.81 N/kg) and h is the height (7m). Please note that sometimes teachers use g = 10, in order to simplify the calculations.For the sake of simplicity, I will use 10. If you want to be exact, use 9.81.

So, plugging in our numbers, we get:Ep = 8 kg * 7 m * 10 N/kg = 560 J

When the toy falls, its energy is transforming from potential to kinetic.So, when the height is 0, all the energy is kinetic.

Ek = 1/2 * m * v^2

So, we know that energy is conserved, which means that the initial energy is the same as the final energy. We had calculated the initial energy to be 560 J, which means that Ek is also 560J

Substituting, we get that:

1/2 * 8 kg * v^2 = 560

V^2 = 140

So, v is the square root of 140, which is 11.83 m/s..