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A rock is dropped from the top of a diving platform into the swimming pool below. Will the distance traveled by the rock in a 0.1-second interval near the top of its flight be the same as the distance covered in a 0.1-second interval just before it hits the water? Explain.

Answer :

Answer:

No the distance traveled in last 0.1 s is not same as that the distance traveled in first 0.1 s

so it will cover more distance in last 0.1 s then the distance in first 0.1 s

Explanation:

As we know that when stone is dropped from the diving board then its velocity at the time of drop is taken to be ZERO

so here we can say that its displacement from the top position in next 0.1 s is given as

[tex]d_1 = v_y t + \frac{1}{2}at^2[/tex]

[tex]d_1 = 0 + \frac{1}{2}(9.81)(0.1)^2[/tex]

[tex]d_1 = 0.05 m[/tex]

Now during last 0.1 s of its motion the stone will attain certain speed

so we will have

[tex]d_2 = v_y(0.1) + \frac{1}{2}(9.81)(0.1)^2[/tex]

[tex]d_2 = 0.1 v_y + 0.05 m[/tex]

so it will cover more distance in last 0.1 s then the distance in first 0.1 s

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