Himadas, a sprinter, runs with constant velocity 3 m/sec through a forest. A fruit from a tree, which is right above him, starts to fall from a height of 20 meters when Himadas starts running. How far behind Himadas will the fruit land?

8 m

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9 m

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6 m

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12 m

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Solution

The correct option is C 6 m
$s = u t + \frac{1}{2} a t^{2}$
$\Rightarrow - h = \frac{1}{2} (- 10) t^{2}$
$\Rightarrow t^{2} = \frac{2 h}{10} = \frac{2 \times 20}{10} = 4$
t = 2 sec
The distance covered by Himadas during this time is,
S = v.t = 3(2) = 6 m

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Himadas, a sprinter, runs with constant velocity 3 m/sec through a forest. A fruit from a tree, which is right above him, starts to fall from a height of 20 meters when Himadas starts running. How far behind Himadas will the fruit land?

The correct option is C 6 m s=ut+12at2 ⇒−h=12(−10)t2 ⇒t2=2h10=2×2010=4 t = 2 sec The distance covered by Himadas during this time is, S = v.t = 3(2) = 6 m

The correct option is C 6 m
$s = u t + \frac{1}{2} a t^{2}$
$\Rightarrow - h = \frac{1}{2} (- 10) t^{2}$
$\Rightarrow t^{2} = \frac{2 h}{10} = \frac{2 \times 20}{10} = 4$
t = 2 sec
The distance covered by Himadas during this time is,
S = v.t = 3(2) = 6 m