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Motion — Science STD 9 — Question

Gujarat BoardEnglish MediumSTD 9ScienceMotion4 Marks
Question
An object is dropped from rest at a height of 150m and simultaneously another object is dropped from rest at a height 100m. What is the difference in their heights after 2s if both the objects drop with same accelerations? How does the difference in heights vary with time?
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Answer

For first object given, u = 0 (because object dropped from rest) and time (t) = 2s

from second equation of motion, the distance covered by first object in 2s is

$\text{h}=\text{ut}+\frac{1}{2}\text{gt}^2$

$\text{h}=0\times2+\frac{1}{2}\times10\times(2)^2\ [\because\text{g}=10\text{m/ s}^{2}]$

$\text{h}=0+\frac{1}{2}\times10\times4=20\text{m}$

Height of first object from the ground after 2s (h1) = 150m - 20m = 130m for second object guven, u = 0 and time (t) = 2s

From second equation of motion, the distance covered by second object in 2s is 

$\text{h}=\text{ut}+\frac{1}{2}\text{gt}^2=0\times2+\frac{1}{2}\times10\times(2)^2\ [\because\text{g}=10\text{m/ s}^2]$

$=0+\frac{1}{2}\times1=\times4=20\text{m}$

Height of second object from the ground after 2s then h2 = 100m – 20m = 80m Now, difference in the height after 2s = h1 – h2 = 130 – 80 = 50 m

The difference in hights of the objects will remain same with time as both the objects have been dropped from rest and are falling with same acceleration i.e (acceleration due to gravity).

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