A bullet of mass 10 gm is fired horizontally with a velocity $1000 m s^{- 1}$ from a rifle situated at a height 50 m above the ground. If the bullet reaches the ground with a velocity $500 m s^{- 1}$ , the work done against air resistance(magnitude) in the trajectory of the bullet in Joule, is $(g = 10 m s^{- 2})$ :

5005

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3755

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3750

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17.5

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Solution

The correct option is D 3755
W.D against air resistance is the energy lost , which is
$W = (m g h + \frac{1}{2} m u^{2}) - \frac{1}{2} m v^{2}$

$= \frac{1}{2} m (u^{2} - v^{2}) + m g h$

$= 10^{- 2} (\frac{1}{2} (1000^{2} - 500^{2}) + 10 \times 50)$

$= 3755 J$

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A bullet of mass 10 gm is fired horizontally with a velocity 1000 ms−1 from a rifle situated at a height 50 m above the ground. If the bullet reaches the ground with a velocity 500 ms−1, the work done against air resistance(magnitude) in the trajectory of the bullet in Joule, is (g=10 ms−2):

The correct option is D 3755W.D against air resistance is the energy lost , which is W=(mgh+12mu2)−12mv2=12m(u2−v2)+mgh=10−2(12(10002−5002)+10×50)=3755J

The correct option is D 3755
W.D against air resistance is the energy lost , which is
$W = (m g h + \frac{1}{2} m u^{2}) - \frac{1}{2} m v^{2}$

$= \frac{1}{2} m (u^{2} - v^{2}) + m g h$

$= 10^{- 2} (\frac{1}{2} (1000^{2} - 500^{2}) + 10 \times 50)$

$= 3755 J$