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

Given:-

• Initial Velocity of stone = 0m/s ( As it was in rest )

• Time taken = 4s.

• Acceleration due to gravity = 9.8m/s

To Find:-

• The Height of Cliff and the velocity with which the stone reaches the ground.

Formulae used:-

• s = ut + ½ × a × t²

• v² - u² = 2as

Where,

• s = Distance
• u = Initial Velocity
• a = Acceleration due to gravity
• t = Time
• v = Final velocity

Now,

→ s = ut +½ × a × t²

→ s = 0 × 4 + ½ × 9.8 × (4)²

→ s = 0 + 4.9 × 16.

→ s = 78.4m

Hence, The Height of the cliff is 78.4m.

Therefore,

→ v² - u² = 2as

→ v² - (0)² = 2 × 9.8 × 78.4

→ v² = 1536.64

→ √v² = √1536.64

→ v = 39.2m/s

Hence, The Final Velocity of stone is 39.2m/s.

Answer 2

Answer:

s= 72 m

v=36 m/s

Explanation:

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t= 4 seconds

g or acceleration= 9m/s

u= 0m/s  <because it starts from rest>

using second equation we get

s=ut +0.5 at²

s=0.5*9*4*4

s= 72 m

using 1 equation

v=u + at

v= 9*4

v=36 m/s

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Aditya