Question

a stone is thrown vertically upward with a speed of 10 m/s from edge of a cliff 65m high .what will be its speed just before hitting the bottom ?

Answer 1

Answer:

37m/s (approx)

Explanation:

Given in the attachment

Note:

formula of height for a body projected upward is H= v²-u²

--------

2g

Answer 2

Answer:

The answer is 37.4.

Explanation:

Speed is a scalar quantity that signifies "how fast an object is moving." Speed can be thought of as the rate at which an object obscures distance. A fast-moving object has an increased speed and surrounds a relatively large distance in a concise amount of time.

Here on the question provided,

$u=10 \mathrm{~m} / \mathrm{s}$

$\mathrm{s}=-6 \mathrm{sm}$

and,

$a=-\mathrm{g}=-10 \mathrm{~m} / \mathrm{s}^{2}$

Applying in the formula,

$\mathrm{v}^{2} &=\mathrm{u}^{2}+2 \mathrm{as}$

$&=(10)^{2}+2(-10)(-65)$

$&=100+1300$

We get,

$&=1400$

$\mathrm{v} &=\sqrt{1400}=10 \sqrt{14} \mathrm{~m} / \mathrm{s}\end{aligned}$

speed just before hitting the bottom

$&=10 \sqrt{14} \mathrm{~m} / \mathrm{s}$

$&=37.4 \mathrm{~m} / \mathrm{s}$

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