Question

a stone is dropped from the top of a building, which is 65m high, with what velocity will it hit the ground​

Answer 1

Answer:

35.69 m/s (approx.)

Explanation:

Given:

Initial velocity (u) = 0 m/s    (as it is dropped from rest)

Displacement (s) = 65 m     (height of the building)

Acceleration (a) = g (acceleration due to gravity) = 9.8 m/s²

To find:

Final velocity (v)

Method to find:

v² = u² + 2as

⇒ v² = 0 + (2*9.8*65)

⇒ v² = 1274

⇒ v = √1274

⇒ v = 35.6931365951 ≈ 35.69 m/s

Hence, the stone will hit the ground with a velocity of roughly 35.69 m/s.

Answer 2

Given:-

→ A stone is dropped from the top of

a buiding.

→ Height of the building = 65m

To find:-

→ Velocity with which the stone will hit

the ground.

Solution:-

In this case :-

• Initial velocity of the stone will be zero .

• Acceleration due to gravity will be +ve

as the stone is freely falling.

By using the 3rd equation of motion, we get :-

v² - u² = 2gh

Where :-

• v is the final [required velocity].

• u is the initial velocity of the stone.

• g is acceleration due to gravity.

• h is height of the building.

Substituting values, we get :-

=> v² - 0 = 2(9.8)65

=> v² = 1274

=> v = √1274

=> v = 35.69 m/s

Thus, the stone will hit the ground with a velocity of 35.69m/s .

Some Extra Information:-

The three equations of motion, for a body freely falling under gravity are :-

• v = u + gt

• h = ut + ½gt²

• v² - u² = 2gh