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

Answer:

• Potential Energy = 9075 J
• Height of building = 151.25 m

Given that a bowling ball of mass 6kg is dropped from the top of a building of height h. It hits the ground with a speed of v = 55 m/s

We have to find the following:

1. How much Potential energy the ball had at the top of the building.
2. How tall was the building?

To Find the Potential energy at the top of the building, we use the formula

⇒ P E = mgh

Where,

• m = mass
• g = acceleration due to gravity
• h = height

In this case,

⇒ P E = 6 × 10 × h [ ∵ g = 10 m/s² ]

⇒ P E = 60h ...(i)

Now, According to the conservation of energy in case of absence of air friction, the potential will be converted into kinetic energy on striking the ground.

So, Let's find Kinetic energy which will be our Potential energy at the top of the building.

⇒ K E = 1/2 mv²

Where,

• m = mass
• v = velocity

⇒ K E = 1/2 × 6 × (55)²

⇒ K E = 3 × 3025

⇒ K E = 9075 J

Hence, 9075 J is the potential energy at the top of the building. Since we have an equation of potential energy at the top of the building of height h. let's substitute the value in eq.(i),

⇒ 60h = 9075

⇒ h = 151.25 m

So, The height of the building is 151.25 m.

We can also solve this question using the formula,

• v = √(2gh)

After getting the height h, we can then find the potential energy. well, this is another way of doing the same thing I did with explanation.

Answer 2

Answer:

• Potential Energy = 9075 J
• Height = 151.25 m

Explanation:

Given:

• Mass of ball (m) = 6 kg
• Velocity (v) = 55 m/s

To Find:

• How tall was the building (Height).
• Potential Energy

Now, we know that

$\implies{\tt{P.E = mgh}}$

$\implies{\tt{P.E = 6\times 10\times h\;\;[Take\;g=10\;m/s^{2}]}}$

$\implies{\tt{P.E = 60h\;\;...(1)}}$

When it is about to reach the ground,  P.E will fully be converted into K.E.

So,

$\implies{\tt{K.E=P.E...\;(2)}}$

$\implies{\tt{K.E=\dfrac{1}{2}\;mv^{2}}}$

$\implies{\tt{K.E=\dfrac{1}{2}\times 6\times 55^{2}}}$

$\implies{\tt{K.E=\dfrac{1}{2}\times 6\times 3025}}$

$\implies{\tt{K.E=\dfrac{1}{2}\times 18150}}$

$\implies{\boxed{\tt{K.E=9075\;J}}}$

Hence, Potential Energy at the top of the building = 9075 J.

Now, Equalate both the equations,

$\implies{\tt{K.E=P.E}}$

$\implies{\tt{9075=60h}}$

$\implies{\tt{\dfrac{9075}{60}=h}}$

$\implies{\boxed{\tt{h=151.25\;m}}}$

Hence, Height of the building = 151.25 m.

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