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Motion in a Straight Line — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Straight Line5 Marks
Question
A perfectly elastic rubber ball is dropped from the top of a building. A man standing in front of a window $2m$ high notes that the ball takes a time of $0.2s$ in crossing the window. The ball strikes the ground suffering a perfectly elastic collision and reappears at the bottom of the window during its upward journey again after $2$ seconds. What is (1) the height of the building and (2) the height of the bottom of the window above the ground? Take $g = 10ms^{-2}$.
✓

Answer

Let A be the top of the building at a height X above the ground. BC is the window of height 2m. Let $t_1$ be the time taken by the ball to fall from A to B.

 Then the time taken to fall through a distance $(x + 2) m$ is $(t_1 + 0.2) s$. Then $\text{x}=\frac{1}{2}\times10\times\text{t}^2_1\ \dots(1)$ and $\text{x}+2=\frac{1}{2}\times10\times(\text{t}_2+0.2)^2$ From Eqns. (1) and (2), we have $5\text{t}^2_1+2=5\text{t}^2_1+0.2+2\text{t}_1$
$1.8=2\text{t}_1$
$\therefore \text{t}_1=0.9\text{s}$
$\therefore \text{x}=\frac{1}{2}\times10\times(0.9)^2=4.05\text{m}$ Let v_1 be the speed acquired by the ball as it reaches the bottom C of the window. Then $\text{v}_1=0 + 10\times1.1 = 11\text{ms}^{-1}$ During the downward journey the ball experiences an acceleration equal to g and strikes the ground with speed V. Since it undergoes a perfectly elastic collision its speed reverses. The time taken in falling from C to ground to back is 2s. Therefore, time taken to fall from C to ground must be 1s. Hence $\text{X}-(\text{x}+2)=11\times1+\frac{1}{2}\times10(1)^2$
$\text{X}-6.05=11+5=16$
$\therefore \text{X}=22.05\text{m}$ The height of the building is 22.05m and the bottom of the window is at height of 16m from the ground.

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