In the air, it had an average speed of 16 \text{m/s}m/start text, m, slash, s, end text. In the water, it had an average speed of 333 \text{

Question

In the air, it had an average speed of 16 \text{m/s}m/start text, m, slash, s, end text. In the water, it had an average speed of 333 \text{m/s}m/start text, m, slash, s, end text before hitting the seabed. The total distance from the top of the cliff to the seabed is 127 meters, and the stone’s entire fall took 12 seconds.

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Ayla 6 days 2021-10-11T14:21:16+00:00 1 Answer 0

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    2021-10-11T14:22:47+00:00

    Answer:

    Incomplete question

    This is the complete question

    A stone falls from the top of a cliff into the ocean.

    In the air, it had an average speed of 16 m/s. In the water, it had an average speed of 3 m/s before hitting the seabed. The total distance from the top of the cliff to the seabed is 127 meters, and the stone’s entire fall took 12 seconds.

    How long did the stone fall in air and how long did it fall in the water?

    Step-by-step explanation:

    The stone speed in air Sa =16m/s

    The stone speed in water Sw=3m/s

    Total distance travelled is 127m

    Total time taken is 12 sec

    Let time in air be ta

    Tune in water be tw

    Therefore

    Total time =ta +tw

    12=ta+tw. ,Equation 1

    ta=12-tw. ,Equation 2

    Also

    Speed is given as

    Speed = distance /time

    Distance =speed × time

    So, total distance = distance traveled in air + distance travelled in water

    127= Sa×ta + Sw×tw

    127=16ta+3tw. , Equation 3

    Substitute equation 2 into. 3

    127=16(12-tw)+3tw

    127=192-16tw+3tw

    127-192=-13tw

    -65=-13tw

    tw=-65/-13

    tw=5 seconds

    Also from equation 2

    ta=12-tw

    ta=12-5

    ta=7 seconds

    The stone spent 7 seconds in air and spent 5 seconds in water before getting to the seabeds

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