MCQ
A ball takes $(t)$ second to fall from a height $\left(H_1\right)$ and $2 t$ second to fall from a height $\left( H _2\right)$. Then $\sqrt{\frac{ H _1}{ H _2}}$ is (consider ideal motion) :
- A2
- B4
- ✓0.5
- D0.25
✓
Answer
Correct option: C.
0.5
(C) 0.5
$\begin{array}{l}H_1=\frac{1}{2} g(t)^2=\frac{1}{2} g t^2 \\ H_2=\frac{1}{2} g(2 t)^2=\frac{1}{2} g(4 t)^2\end{array}$
$\begin{array}{l}\frac{ H _1}{ H _2}=\frac{\frac{1}{2} g t ^2}{\frac{1}{2} g(4 t )^2}=0.25 \\ \sqrt{\frac{ H _1}{ H _2}}=0.5\end{array}$
$\begin{array}{l}H_1=\frac{1}{2} g(t)^2=\frac{1}{2} g t^2 \\ H_2=\frac{1}{2} g(2 t)^2=\frac{1}{2} g(4 t)^2\end{array}$
$\begin{array}{l}\frac{ H _1}{ H _2}=\frac{\frac{1}{2} g t ^2}{\frac{1}{2} g(4 t )^2}=0.25 \\ \sqrt{\frac{ H _1}{ H _2}}=0.5\end{array}$
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