Question

A stone is dropped from the top of a building which is 180 meters high. If the stone hits the ground after 6 seconds, what will its final velocity be?​

Answer 1

Answer:

Let g = -9.80665m/s^2, t = 6, h = The height to ground distance of the building.

h = - g * t^2/2 = -9.80665 * 18 = -176.519698m.

The stone therefore falls 176.519698m to ground.

Let g = -32.174ft./s^2, t = 6, h = The height to ground distance of the building.

h = - g * t^2/2 = -32.174 * 18 = -579.132000ft.

Explanation:

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Answer 2

Answer :

Height of building = 180m

Time taken by stone to reach at ground = 6s

We have to find final velocity at which stone touch to the ground$.$

★ For a body balling freely under the action of gravity, g is taken positive.

Since constant acceleration acts on stone throughout the motion, we can apply equation of kinematics to solve this problem.

First equation of motion is given by

• v = u + gt

v denotes final velocity

u denotes initial velocity

g denotes gravitational acceleration

t denotes time

⭆ v = u + gt

[For free falling motion, u = 0]

⭆ v = 0 + (10)(6)

⭆ v = 60m/s

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