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A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an initial speed of 52.92 m/s. They hit the ground at the same time. How long does it take the first stone to hit the ground? The acceleration of gravity is 9.8 m/s 2 .

Respuesta :

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

[tex]h=\frac{gt^2}{2}[/tex]

For second stone

[tex]h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}[/tex]---2

Equating 1 &2 we get

[tex]\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}[/tex]

[tex]\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0[/tex]

[tex]13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0[/tex]

[tex]26.46t-35.721-52.92t+142.884=0[/tex]

[tex]t=4.05 s[/tex]

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