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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Salim Ahmed Kawsar · Answer 1 · Feb 05, 2023 15:29:48

According to the equation of motion under gravity v² − u² = 2gs

Where,

u = Initial velocity of the stone = 0 m/s

v = Final velocity of the stone

s = Height of the stone = 19.6 m

g = Acceleration due to gravity = 9.8 ms⁻²

∴ v ² − 0² = 2 × 9.8 × 19.6

⇒ v ² = 2 × 9.8 × 19.6 = (19.6)²

⇒ v = 19.6 ms⁻¹

Hence, the velocity of the stone just before touching the ground is 19.6 ms⁻¹ .

Salim Ahmed Kawsar · Answer 2 · Feb 05, 2023 15:29:48

h = 19.6 m.

Initial velocity u = 0 (∵ It starts from rest)

From equation of motion.

v² = u² + 2gh

v² = 0 + 2 x 9.8 x 19.6

= 19.6 x 19.6

∴ v = \(\sqrt{19.6\times 19.6}\)

= 19.6 ms^-1

Final velocity of stone before touching the ground = 19.6 ms^-1.

2 Answers