Question

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

Answer 1

Answer:

Initial Velocity u=0

Fianl velocity v=?

Height, s=19.6m

By third equation of motion

v

2

=u

2

+2gs

v

2

=0+2×9.8×19.6

v

2

=384.16

⇒v=19.6m/s

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Answer 2

Answer:

velocity just before touching the ground will be 19.6 Meters per second

Solving the Problem:

Finding required from the question:

Initial velocity (u) = 0

Acceleration due to Gravity (g) = 9.8 m/s²

Height (h) = 19.6 meters

Final Velocity (v) = ?

Formula:

v² = u² + 2 g x h

Adding the numbers to the Formula:

v² = (0)² + 2 × 9.8 × 19.6

Solving:

v² = (0)² + 2 × 9.8 × 19.6

v² = 19.6 × 19.6

v² = (19.6)²

v = 19.6 m/s (taking the square away cause we only need 'v')

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