Physics, asked by jaipaldassjethani21, 9 months ago

A stone is thrown vertically upward with velocity of 300m/sec If acceleration due to gravity is 10m/sec what is distance travelled by particle during first second of motion​

Answers

Answered by Anonymous
0

Explanation:

ANSWER BY ADI5166

Initial Velocity u=40

Fianl velocity v=0

Height, s=?

By third equation of motion

v

2

−u

2

=2gs

0−40

2

=−2×10×s

s=

20

160

⇒s=80m/s

Toatl distance travelled by stone = upward distance + downwars distance =2×s=160m

Total Diaplacement =0, Since the initial and final point is same.

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