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Standard XII

Physics

Motion Under Gravity

The maximum h...

Question

The maximum horizontal range of a ball projected with a velocity of $40 m / s$ is (take $g = 9.8 m / s^{2}$ )

157 m

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127 m

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163 m

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153 m

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Solution

The correct option is C $163 m$
Maximum horizontal range $= \frac{u^{2}}{g}$
$= \frac{{(40)}^{2}}{9.8}$ $[∵ u = 40 m / s]$

$= \frac{1600}{9.8}$

$= 163.3 = 163 m$ (approx.)

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The maximum horizontal range of a ball projected with a velocity of 40m/s is (take g=9.8m/s2)

The correct option is C 163mMaximum horizontal range =u2g =(40)29.8 [∵u=40m/s] =16009.8 =163.3=163m (approx.)

The maximum horizontal range of a ball projected with a velocity of $40 m / s$ is (take $g = 9.8 m / s^{2}$ )

The correct option is C $163 m$
Maximum horizontal range $= \frac{u^{2}}{g}$
$= \frac{{(40)}^{2}}{9.8}$ $[∵ u = 40 m / s]$

$= \frac{1600}{9.8}$

$= 163.3 = 163 m$ (approx.)