(A) For maximum range,the angle of projection is $\theta = 45^\circ$. The range $R$ is given as $100\, km = 10^5\, m$.The formula for range is $R = \f

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(A) For maximum range,the angle of projection is $\theta = 45^\circ$. The range $R$ is given as $100\, km = 10^5\, m$.The formula for range is $R = \frac{u^2 \sin(2\theta)}{g}$. Since $\sin(90^\circ) = 1$,we have $R = \frac{u^2}{g}$,so $u^2 = Rg$.The formula for maximum height is $H = \frac{u^2 \sin^2 \theta}{2g}$.Substituting $u^2 = Rg$ and $\theta = 45^\circ$ $(\sin^2 45^\circ = 1/2)$:$H = \frac{(Rg) \times (1/2)}{2g} = \frac{R}{4}$.Given $R = 10^5\, m$,we get $H = \frac{10^5}{4} = 2.5 \times 10^4\, m$.

Class 11 Physics 3-2.Motion in Plane Horizontal Projectile Motion

Medium

$A$ missile is fired for maximum range at your town from a place $100\, km$ away from you. If the missile is fired for maximum range,what is the maximum height reached by the missile? (Take $g = 10\, m/s^2$)

A
$2.5\times10^4\, m$
B
$5.0\times10^4\, m$
C
$7.5\times10^4\, m$
D
$10^5\, m$

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3-2.Motion in Plane

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Horizontal Projectile Motion

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$A$ missile is fired for maximum range at your town from a place $100\, km$ away from you. If the missile is fired for maximum range,what is the maximum height reached by the missile? (Take $g = 10\, m/s^2$)

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