Question

A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

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Answer 1

Answer:

Since the stone is released from the top

Initial velocity  = u = 0m/s

distance travel = 19.6

v²-u²=2 as

v²-(0)=2 x 9.8 x 19.6

v² = 19.6²

v = 19.6 m/s

∴ final velocity of the ball = 19.6 m/s

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Answer 2

Given:-

• Height,h = 19.6 m

• Initial velocity,u = 0 m/s

• Acceleration due to gravity,g = 9.8 m/s²

To be calculated:-

Calculate its final velocity just before touching the ground.

Formula used:-

v² = u² + 2gh

Solution:-

From the third equation of motion for freely falling body;

v² = u² + 2gh

★ Substituting the values in the above formula,we get:

v² = ( 0 )² + 2 × 9.8 × 19.6

v² = 0 + 19.6 × 19.6

v² = ( 19.6 )²

v = 19.6 m/s

Thus,the final velocity of a stone just before touching the ground is 19.6 metres per seconds .

Additional information:-

Velocity : Distance travelled by a body per unit time in given direction is called velocity.

Acceleration : Rate of increase in velocity.

Displacement : The short path between initial position and final position is called Displacement.

Distance : Actual path covered by a body irrespective of its direction.