Question

an object throws upward with a velocity of 40 km/hr and has acceleration 10ms/s² find the maximum height attained by the object?​

Answer 1

Answer:

Initial Velocity u=40

Fianl velocity v=0

Height, s=?

By third equation of motion

v2−u2=2gs

0−402=−2×10×s

s=20160

⇒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.

Answer 2

Answer:

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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