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An object is dropped from 600 m. If it is initially at rest and achieves a top speed of 45 m/s just as it hits the ground, what is its acceleration?​

Sagot :

shreetikabarnwal

Answer:

1.68m/s^2

Explanation:

using V^2=U^2+2×a×s formula.

If,

(Final Velocity)V=45

(Initial Velocity)U=0 because it starts from rest

(Distance)S=600m

(acceleration)a= ?

now,

using formula,

45^2=0^2+2×a×600

2025= 1200a

a=2025÷1200

a = 1.68m/s^2(The unit of acceleration is m/s^2)

Therefore the acceleration is 1.68m/s^2

Hope it works !!!

Answer:

  1. i don't know answer
  2. sorry
  3. mistake
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