# suppose you drop a ball from a

suppose you drop a ball from a certainheightand find itsfinalspeedis 20 m/s when it hits the ground.if you throw theobjectvertically downwards from the sameheight with an initialspeed of 20 m/s, will its finalspeed be 40m/s? carefully explainwhy or why not (ignore friction)

Answer:

In the case of free fall , initial speed of an object will bezero.

Initial speed of the ball be u = 0

Then final speed v ( speed of the ball when it hits the ground )can be calculated when the height is given using the formula

v^{2} = 2 g h

where g = 9.81 m/s^{2} is the acceleration due togravity

h is the height from which ball is falling.

Here final speed is given and we have to calculate theheight.

Final speed v = 20 m/s

( 20 )^{2} = 2 × 9.81 × h

400 = 19.62 h

h = 400 / 19.62

h = 20.38 m

So ball is released from a height h = 20.38 m

We can use this data to calculate the final speed of an objectwhen it is thrown with an initial speed u’ from the sameheight.

Initial speed of the ball u’ = 20 m/s

Final speed v’ is given by

v’^{2} – u’^{2} = 2 g h

v’^{2} – ( 20 )^{2} = 2 × 9.81 × 20.38

v’^{2} – 400 = 400

v’^{2} = 400 + 400

v’^{2} = 800

v’ = 800

v’ = 28.28 m/s

When the ball is thrown downwards with an initial speed u’ = 20m/s , its final speed would be 28.28 m/s

It is not 40 m/s . Final speed does not linearly depend onheight ( provided the g value is constant.) It depends on thesquare root of the height from which object is falling.

When an object is freely falling , final speed represents thechange in speed of the object ( from 0 to a specific value )

Here final speed depends on root of the sum of square of initialspeed and change in the speed.

It is not a direct sum of initial speed and change in the speedduring the fall.

That’s why we got the final speed of 28.28 m/s and not40m/s.