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# An object is moving with a velocity of 5ms-1. At what height must a similar body...

### Question

An object is moving with a velocity of 5ms-1. At what height must a similar body be situated to have a potential energy equal in value with the kinetic energy of the moving body? [g ≈ 10ms-2]

A) 1.3m

B) 25.0m

C) 1.0m

D) 20.0m

### Explanation:

Let's start from K.E = P.E:

$$\text{mgh} = \frac{1}{2}\text{mv}^2$$

$$\text{mgh} = \cfrac{\text{mv}^2}{2}$$

Cross multiply:

It gives us:

$$2\text{mgh} = \text{mv}^2$$

Divide both sides by m:

It gives:

$$2\text{gh} = \text{v}^2$$

To get the value of h, divide both sides by 2g

The formula to use will become:

$$\text{h} = \cfrac{\text{v}^2}{2\text{g}}$$

Then $$\text{v} = 5\text{ms}^{-1}$$

g = 10. Put this into the above formula:

$$\text{h} = \cfrac{5^2}{2 \times 10}$$

$$\text{h} = \cfrac{25}{20}$$

$$\text{h} = 1.3\text{m}$$

Explanation provided by Idowu Taye Ebenezer

## Dicussion (1)

• Let's start from K.E = P.E:

$$\text{mgh} = \frac{1}{2}\text{mv}^2$$

$$\text{mgh} = \cfrac{\text{mv}^2}{2}$$

Cross multiply:

It gives us:

$$2\text{mgh} = \text{mv}^2$$

Divide both sides by m:

It gives:

$$2\text{gh} = \text{v}^2$$

To get the value of h, divide both sides by 2g

The formula to use will become:

$$\text{h} = \cfrac{\text{v}^2}{2\text{g}}$$

Then $$\text{v} = 5\text{ms}^{-1}$$

g = 10. Put this into the above formula:

$$\text{h} = \cfrac{5^2}{2 \times 10}$$

$$\text{h} = \cfrac{25}{20}$$

$$\text{h} = 1.3\text{m}$$