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Standard XII

Physics

Range

In between tw...

Question

In between two hills of heights $100 m$ and $92 m$ , there is a valley of breadth $16 m$ . If a vehicle jumps from the first hill to the second one, the approximate minimum speed of the vehicle is (assume $g = 9.8 m / s^{2}$ )

16 m / s

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12 m / s

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9 m / s

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10 m / s

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Solution

The correct option is A $12 m / s$

$D =$ Total horizontal distance $= 16 m$
Time of flight, $t_{t} = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 (100 - 92)}{9.8}} = \frac{4}{\sqrt{g}}$
So, $D = v t_{t} = 16 = v \times \frac{4}{\sqrt{g}}$
$v = 4 \sqrt{g} = 12 m / s$

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In between two hills of heights 100m and 92m, there is a valley of breadth 16m. If a vehicle jumps from the first hill to the second one, the approximate minimum speed of the vehicle is (assume g=9.8m/s2)

The correct option is A 12m/s D= Total horizontal distance =16mTime of flight, tt=√2hg=√2(100−92)9.8=4√gSo, D=vtt=16=v×4√gv=4√g=12 m/s

In between two hills of heights $100 m$ and $92 m$ , there is a valley of breadth $16 m$ . If a vehicle jumps from the first hill to the second one, the approximate minimum speed of the vehicle is (assume $g = 9.8 m / s^{2}$ )

The correct option is A $12 m / s$

$D =$ Total horizontal distance $= 16 m$
Time of flight, $t_{t} = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 (100 - 92)}{9.8}} = \frac{4}{\sqrt{g}}$
So, $D = v t_{t} = 16 = v \times \frac{4}{\sqrt{g}}$
$v = 4 \sqrt{g} = 12 m / s$