TG EAPCET 2025 (Online) 2nd May Evening Shift Physics - Motion in a Straight Line Question 28 English Explanation 1

As object return with smaller speed that means there is a loss of energy. This loss of energy occurs due to viscous drag of atmosphere. While object is moving up let viscous drag produces an deceleration ' $a$ '.

Then,

$$ \begin{array}{ll} & v^2-u^2=2 a s \\ \Rightarrow \quad & 0-(10)^2=2(-(g+a)) h \\ \text { or } \quad & 100=2(g+a) h \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $$

While coming down,

TG EAPCET 2025 (Online) 2nd May Evening Shift Physics - Motion in a Straight Line Question 28 English Explanation 2

$$ \begin{aligned} &\begin{array}{ll} \Rightarrow & 64-0=2(-(g-a))(-h) \\ \Rightarrow & 64=2(g-a) h \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array}\\ &\text { Dividing (i) and (ii) we get }\\ &\begin{aligned} & \frac{100}{64}=\frac{g+a}{g-a} \\ \Rightarrow \quad & 100 g-100 a=64 g+64 a \\ & 36 g=164 a \text { or } a=\frac{36}{164} g \end{aligned} \end{aligned} $$

Substituting ' $a$ ' in either (i) or (ii) we get ' $h$.

From (ii)

$$ \begin{aligned} & & 64 & =2\left(g-\frac{36}{164} g\right) h \\ \Rightarrow & & 32 & =\left(\frac{164-36}{164}\right) g h \Rightarrow \\ \Rightarrow & & h & =4.1 \mathrm{~m} \end{aligned} $$

Motion In A Straight Line

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