Question

a ball is droped from a height of 176.4m. how much time will take to reach the ground​

Answer 1

Answer:

You must know the speed of the ball

Step-by-step explanation:

Answer 2

Answer:

Q.Here Is Your Answer .

• 176.4= 9.8t^2/2
• 176.4= 9.8t^2/218= t^2/2
• 176.4= 9.8t^2/218= t^2/236= t^2
• 176.4= 9.8t^2/218= t^2/236= t^2t= 6sec for the first ball to touch the water surface when released from the 176.4 m height.
• 176.4= 9.8t^2/218= t^2/236= t^2t= 6sec for the first ball to touch the water surface when released from the 176.4 m height.2nd ball has to take 4sec to reach the water surface
• 176.4= 9.8t^2/218= t^2/236= t^2t= 6sec for the first ball to touch the water surface when released from the 176.4 m height.2nd ball has to take 4sec to reach the water surface176.4= u4+9.8*16/2
• 176.4= 9.8t^2/218= t^2/236= t^2t= 6sec for the first ball to touch the water surface when released from the 176.4 m height.2nd ball has to take 4sec to reach the water surface176.4= u4+9.8*16/2176.4–78.4= u4
• 176.4= 9.8t^2/218= t^2/236= t^2t= 6sec for the first ball to touch the water surface when released from the 176.4 m height.2nd ball has to take 4sec to reach the water surface176.4= u4+9.8*16/2176.4–78.4= u498= 4u
• 176.4= 9.8t^2/218= t^2/236= t^2t= 6sec for the first ball to touch the water surface when released from the 176.4 m height.2nd ball has to take 4sec to reach the water surface176.4= u4+9.8*16/2176.4–78.4= u498= 4uu=24.5.sec.

• Step-by-step explanation:

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