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Standard XII

Physics

Defining Kinetic Energy

A ball is thr...

Question

A ball is thrown vertically upwards with a velocity of $10 m/s$ . It returns to the ground with velocity of $9 m/s$ . If $g = 9.8 {m/s}^{2}$ , then the maximum height attained by the ball is nearly (assume air resistance to be uniform)

5.1 m

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4.1 m

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4.61 m

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5.0 m

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Solution

The correct option is C $4.61 m$
Total energy loss due to air friction $= K_{i} - K_{f} = \frac{1}{2} m (100 - 81)$
$= \frac{19}{2} m (m = mass)$
Half of the loss will be in upward journey.
Hence, $m g h + \frac{1}{2} (\frac{19}{2} m) = K_{i} = \frac{1}{2} m (100) = 50 m$
$∴ h = 4.61 m$

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A ball is thrown vertically upwards with a velocity of 10 m/s. It returns to the ground with velocity of 9 m/s. If g=9.8 m/s2, then the maximum height attained by the ball is nearly (assume air resistance to be uniform)

The correct option is C 4.61 mTotal energy loss due to air friction =Ki−Kf=12m(100−81) =192m (m=mass) Half of the loss will be in upward journey. Hence, mgh+12(192m)=Ki=12m(100)=50m ∴h=4.61 m

A ball is thrown vertically upwards with a velocity of $10 m/s$ . It returns to the ground with velocity of $9 m/s$ . If $g = 9.8 {m/s}^{2}$ , then the maximum height attained by the ball is nearly (assume air resistance to be uniform)

The correct option is C $4.61 m$
Total energy loss due to air friction $= K_{i} - K_{f} = \frac{1}{2} m (100 - 81)$
$= \frac{19}{2} m (m = mass)$
Half of the loss will be in upward journey.
Hence, $m g h + \frac{1}{2} (\frac{19}{2} m) = K_{i} = \frac{1}{2} m (100) = 50 m$
$∴ h = 4.61 m$