The maximum height reached for new angle of projection is

\frac{2 A^{2}}{B}

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\frac{A^{2}}{2 B}

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\frac{A^{2}}{4 B}

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\frac{2 A^{2}}{3 B}

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Solution

The correct option is D $\frac{A^{2}}{4 B}$
$y = A t - B t^{2}$
At maximum height vertical component of velocity is zero, $\frac{d y}{d t} = 0 o r A - 2 B t = 0 o r t = \frac{A}{2 B}$
Therefore, at maximum height: $y = H = A (\frac{A}{2 B}) - B {(\frac{A}{2 B})}^{2} = \frac{A^{2}}{4 B}$

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The maximum height reached for new angle of projection is

The correct option is D A24By=At−Bt2At maximum height vertical component of velocity is zero, dydt=0orA−2Bt=0ort=A2B Therefore, at maximum height: y=H=A(A2B)−B(A2B)2=A24B

The correct option is D $\frac{A^{2}}{4 B}$
$y = A t - B t^{2}$
At maximum height vertical component of velocity is zero, $\frac{d y}{d t} = 0 o r A - 2 B t = 0 o r t = \frac{A}{2 B}$
Therefore, at maximum height: $y = H = A (\frac{A}{2 B}) - B {(\frac{A}{2 B})}^{2} = \frac{A^{2}}{4 B}$