**Downward motion**

So, the stone goes up for 2 sec. The distance covered in these 2 sec can be found by

2gs = v_{f}^{2} – v_{i}^{2}. Put the values,

2(-9.8)s = 0 – (19.6)^{2} Or s = (19.6 * 19.6)/(2 * 9.8) Or s = 384.16/19.6 = 19.6 m

So, the total distance the stone drops through is (156.8 + 19.6) = 176.4 m under the action of gravity. Therefore, to find time t, we apply, s = v_{i}t + ½ at^{2}, where v_{i} = 0. Put the values,

176.4 = 0 + ½ × 9.8 × t^{2} Or 176.4 = 4.9t^{2} Or t^{2} = (176.4)/4.9 = t^{2} = 36 Or t = **6 s** … (2)

What if we solve this problem simply by putting values in the equation “time= distance / speed”?

SanaFormula t = s/v is applicable only in the case when a body is moving with uniform velocity. Here the stone is continuously under the influence of gravity and the velocity is changing every moment.

Nice

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