First slide

Displacement in nth sec

Question

A stone falls freely under gravity. It covers distances $h_{1}, h_{2} and h_{3}$ in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between $h_{1}, h_{2} and h_{3}$ is

Moderate

A

$h_{2} = 3 h_{1} a n d h_{3} = 3 h_{2}$
B

$h_{1} = h_{2} = h_{3}$
C

$h_{1} = 2 h_{2} = 3 h_{3}$
D

$h_{1} = \frac{h_{2}}{3} = \frac{h_{3}}{5}$

Solution

Distance covered by the stone in first 5 seconds
(i.e. t = 5s) is $h_{1} = \frac{1}{2} g {(5)}^{2} = \frac{25}{2} g - - - (i)$
Distance travelled by the stone in next 5 seconds (i.e. t = 10 s) is
$h_{1} + h_{2} = \frac{1}{2} g {(10)}^{2} = \frac{100}{2} g - - - (i i)$
Distance travelled by the stone in next 5 seconds (i.e. t =15 s) is
$h_{1} + h_{2} + h_{3} = \frac{1}{2} g {(15)}^{2} = \frac{225}{2} g - - - (i i i)$
Subtract (i) from (ii), we get
$(h_{1} + h_{2}) - h_{1} = \frac{100}{2} g - \frac{25}{2} g = \frac{75}{2} g h_{2} = \frac{75}{2} g = 3 h_{1} - - - (i v)$
Subtract (ii) from (iii), we get
$(h_{1} + h_{2} + h_{3}) - (h_{2} + h_{1}) = \frac{225}{2} g - \frac{100}{2} g h_{3} = \frac{125}{2} g = 5 h_{1} - - - (v)$
From (i), (iv) and (v), we get
$h_{1} = \frac{h_{2}}{3} = \frac{h_{3}}{5}$

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