Question

If an object is thrown from a cliff of unknown height, it’s final velocity is 100 m/s. What is the height of the cliff?

Answer 1

Answer:

PE = KE

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

$gh = \frac{1}{2} v {}^{2}$

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

$h = \frac{v {}^{2} }{2g}$

$h = \frac{100^{2} }{2 \times9.807 }$

$h = \frac{10000}{19.614 }$

$h = 509.83991 \: m$

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Answer 2

The height of the cliff is 500 meters

Explanation:

Given as :

An object is thrown from a cliff of unknown height .

The final velocity of the object = v = 100 m/s

Let The height of the cliff = h meters

Let The initial velocity of the object = u m/s

According to question

Initial the object is at rest

So, initial velocity of object = u = 0 m/s

Now,

For object

kinetic energy of object = potential energy of object

i.e   KE = PE

Or, $\dfrac{1}{2}$ × mass × velocity²  = mass × acceleration due to gravity × height

Or,  $\dfrac{1}{2}$ × m × v²  = m × g × h

Or, ,  $\dfrac{1}{2}$ × (100 m/s)²  = 10 m/s² × h

Or, 10 m/s² × h = 5000  m²/s²

∴                    h = $\dfrac{5000}{10}$

i.e                  h = 500 meters

So, The height of the cliff = h = 500 m

Hence, The height of the cliff is 500 meters   Answer