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An object is dropped from rest at a height of $150\, m$ and simultaneously another object is dropped from rest at a height $100 \,m$. What is the difference in their heights after $2\,\sec $ if both the objects drop with same accelerations ? How does the difference in heights vary with time ?
Solution
Initial difference in height $=(150-100) \,m =50 \,m$
Distance travelled by first body in $2 \,s=h_{1}=0+\frac{1}{2}\, g(2)^{2}=2\, g$
Distance travelled by another body in $2 \,s=h_{2}=0+\frac{1}{2} \,g(2)^{2}=2 \,g$
After $2 \,s$, height at which the first body will be $= h _{1}^{\prime}=150-2\, g$
After $2\, s,$ height at which the second body will be $= h _{2}^{\prime}=100-2\, g$
Thus, after $2\, s$, difference in height $=150-2\, g -(100-2\, g )$
$=50 \,m =$ initial difference in height
Thus, difference in height does not vary with time.
Similar Questions
The following table show os the positon of three persons between $8.00\, am$ to $8.20\, am$.
| Time | Position (in $km$) | ||
| Person $A$ | Person $B$ | Person $C$ | |
| $8.00 \,am$ | $0$ | $0$ | $0$ |
| $8.05 \,am$ | $4$ | $5$ | $10$ |
| $8.10\, am$ | $13$ | $10$ | $19$ |
| $8.15 \,am$ | $20$ | $15$ | $24$ |
| $8.20\, am$ | $25$ | $20$ | $27$ |
$(i)$ Who is moving with constant speed ?
$(ii)$ Who has travelled maximum distance between $8.00\, am$ to $8.05\, am$ ?
$(iii)$ Calculate the average speed of person $'A^{\prime}$ in $k m h^{-1}$
