MCQ
An object thrown vertically up from the ground passes the height $5 m$ twice in an interval of $10 s$. time of flight is ........ $s$
- A$\sqrt{28}$
- B$\sqrt{86}$
- ✓$\sqrt{104}$
- D$\sqrt{72}$
✓
Answer
Correct option: C.
$\sqrt{104}$
c
(c)
(c)
$h=5\, m \text { (given) }$
$t_2-t_1=10 \,s$
$T \rightarrow$ Time taken to reach the highest point.
$t_1=T-\sqrt{T^2-\frac{2 h}{g}}, t_2=T+\sqrt{T^2-\frac{2 h}{g}}$
$t_2-t_1=T+\sqrt{T^2-\frac{2 h}{g}}-T+\sqrt{T^2-\frac{2 h}{g}}$
$\Rightarrow 10=2 \sqrt{T^2-\frac{2 \times 5}{10}}$
$\Rightarrow 5=\sqrt{T^2-1} \Rightarrow 25=T^2-1$
$T^2=26$
$\Rightarrow T=\sqrt{26}$
$\text { Total time of flight } \Rightarrow 2 T=2 \sqrt{26}=\sqrt{4 \times 26}=\sqrt{104} \,s$
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