A ball is thrown vertically upwards. It has a speed of $10 m / s e c$ , when it has reached one half of its maximum height. How high does the ball rise? Take $g = 10 m / s^{2}$ )

10 m

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5 m

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15 m

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20 m

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Solution

The correct option is A $10 m$
Maximum height = $h = \frac{u^{2}}{2 g}$ .......1
, $u$ being the initial velocity
using third equation $v^{2} - u^{2} = 2 g \times h e i g h t$ $g a i n e d$ $= 2 g \times \frac{h}{2} = g h$
putting $u^{2} = 2 g h$ in above equation from first equation we get $v^{2} - 2 g h = g h$ or $h = \frac{v^{2}}{3 g} = \frac{10 \times 10}{10} = 10 m e t e r$
Option A

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A ball is thrown vertically upwards. It has a speed of 10m/sec, when it has reached one half of its maximum height. How high does the ball rise? Take g=10m/s2)

The correct option is A 10mMaximum height = h=u22g.......1 , u being the initial velocityusing third equation v2−u2=2g×height gained =2g×h2=ghputting u2=2gh in above equation from first equation we get v2−2gh=gh or h=v23g=10×1010=10meterOption A

A ball is thrown vertically upwards. It has a speed of $10 m / s e c$ , when it has reached one half of its maximum height. How high does the ball rise? Take $g = 10 m / s^{2}$ )