A ball is dropped from height h. the velocity attain by ball just befo

1.	A ball is dropped from height h. the velocity attain by ball just before reaching the ground is
Answer» A ball is DROPPED from height h. The ball has no velocity at start. So, Initial velocity = 0 m/s The ball is accelerated by Earth's Gravity. Acceleration of the ball = g. Now, The ball comes down a height. Acceleration due to gravity is 9.8 m/s² that means, the body will be change its velocity by 9.8 m/s every second during it's fall. Equations of motion v = u + at s = UT + at²/2 s = ut + at²/2 v² - u² = 2as We have the initial velocity of the ball, height it travels. So, Equation 1, 2 can't be used as we don't know the time it falls. Therefore, THIRD Equation is to be used. v² = u² + 2as v² = 0² + 2gh v² = 2gh $v = \sqrt{2gh}$ Therefore, The velocity of the ball is $v = \sqrt{2gh}$

A ball is dropped from height h. the velocity attain by ball just before reaching the ground is

Answer»

A ball is DROPPED from height h.

The ball has no velocity at start.

So, Initial velocity = 0 m/s

The ball is accelerated by Earth's Gravity.

Acceleration of the ball = g.

Now, The ball comes down a height.

Acceleration due to gravity is 9.8 m/s² that means, the body will be change its velocity by 9.8 m/s every second during it's fall.

Equations of motion

v = u + at

s = UT + at²/2

s = ut + at²/2 v² - u² = 2as

We have the initial velocity of the ball, height it travels. So, Equation 1, 2 can't be used as we don't know the time it falls. Therefore, THIRD Equation is to be used.

v² = u² + 2as

v² = 0² + 2gh

v² = 2gh

$v = \sqrt{2gh}$

Therefore, The velocity of the ball is

$v = \sqrt{2gh}$

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