Question

A stone dropped from the top of a cliff reaches its foot in 4seconds claculate height of the cliff and the velocity with which the stone reaches the ground (g=9m/s)​

Answer 1

Answer:

Height of Cliff = 72m , Final Velocity = 36 m/s

Explanation:

Stone dropped from state of rest,

initial velocity (u) = 0 m/s,

Time taken (t) = 4,

Final Velocity (v) = ?,

Displacement (s) = ? ,

Acceleration (a) = 9 m/s² {Given}

From Kinematics (Formula of Final Velocity) which is

v = u + at

v = 0 + 9 × 4

v = 36 m/s ----(1)

From Kinematics (Formula of Displacement) which is

v²-u² = 2as

v²-u² / 2a = s

(36) ² - (0) ² /2(9) = s

1296/18 = s

s = 72m -----(2)

Hence, The height of the cliff was 72 m, and the velocity of Stone before hitting the ground was 36 m/s.

Answer 2

Answer:

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