Question

Example 17.12. A stone is dropped from the top of a building, which is 65 m high. With what
velocity will it hit the ground ?​

Answer 1

Explanation:

given that,

A stone is dropped from the top of a building 65 m

Here,

initial velocity of the stone = 0 m/s

[stone was at rest]

gravitational acceleration = -9.8 m/s²

let the time taken by the stone to reach the ground be t

here,

we have,

initial velocity(u) = 0 m/s

gravitational acceleration(g) = -9.8 m/s²

height(h) = -65 m

[height is downwards]

time taken(t) = t

by the gravitational equation of motion,

h = ut + ½ gt²

putting the values,

-65 = 0(t) + ½ × (-9.8)t²

-4.9t = -65

t = -65/-4.9

t = 13.3

so,

time taken by stone to reach ground

➪ 13.3 seconds

now,

let the velocity by which it hit the ground be v

v = u + gt

v = 0 + (-9.8)(13.3)

v = -130.04 m/s

negation of velocity shows the velocity downwards

so,

velocity by which it hit the ground➪ 130.04 m/s

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