Question

A stone is dropped from a certain height and another stone is dropped from the same height after 2 s. What
will be their separation after 10 more seconds?
(1) 115.6 m
(2) 156.5 m
(3) 172.3 m
(4) 215.6 m​

Answer 1

Answer:

Use newtons second law of motion

distance travelled after t seconds

$s (t)=(0)t - \frac{1}{2} g {t}^{2} = 5 {t}^{2}$

Now we need

$s(12) - s(10) \\ = 5( {12)}^{2} - 5(10 {}^{2} ) \\ = 5(144 - 100) \\ = 220 \: m$

**Use g = 9.8 for exact calculation

$4.9 \times 44 = 215.6$

Answer 2

Answer:

The separation after 10 more seconds is = 215.6m

Explanation:

When a body falls under gravity when its initial velocity is zero and acceleration is due to gravity and t remains fairly constant when height h is not large, the equation reduces to the form h= 1/2gt²

let us take two stones A and B and our acceleration due to gravity is =9.8m/s

distance covered by stone A in 12 seconds =  1/2gt²

S(A) = 1/2×9.8×144

S(A) = 705.6m

distance covered by stone B IN 10 seconds = 1/2gt²

S(B) = 1/2×9.8×100

S(B) = 490m

after 10 more seconds:

S(A) - S(B) = 705.6 - 490

                 = 215.6m

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