Question

A ball is dropped from a height of 50m. What will be its velocity before touching ground?​

Answer 1

Given :

A ball is dropped from a height of 50 m

To find :

the velocity of the ball just before it reaches the ground

Solution :

As the ball is dropped, the initial velocity of the ball is 0 m/s

The distance the ball has travelled from height of 50 m to the ground is 50 m

S = 50 m

Using third equation of motion,

v² - u² = 2as

where ,

v denotes final velocity

u denotes initial velocity

a denotes acceleration of the object

s denotes the distance covered by the object

In this case, a = g = 10 m/s²

Substitute the values,

v² - 0² = 2(10)(50)

v² = 1000

v = √1000

v = 10√10 m/s

Therefore, the velocity of the ball just before it reaches the ground is 10√10m/s

We calculated the velocity of the ball without using it's mass as in the expression we used, there is no term of mass. So, we do not require the value of mass to find the velocity.

For more details : https://brainly.in/question/34106626

Answer 2

Answer:

$10\sqrt{10}$m/s is the velocity before touch the ground.

Explanation:

Given, a ball is dropped from a height  of 50 m.

So, the initial velocity of the(u)  = 0m/s.

And distance(s) = 50m

let v be the final velocity.

And acceleration due to gravity (g) = 10 m/$s^2$ = a (acceleration)

Now, as we know that, the third law of motion ;

$v^2 - u^2 = 2as$

On putting all the values in the above  equation we get,

⇒$v^2 - 0 = 2 (10)(50)$

⇒v = $\sqrt{1000}$ = $10\sqrt{10}$m/s.

Hence, its velocity before touching the  ground is $10\sqrt{10}m/s$.

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