A particle thrown at angle $45^{\circ}$ with horizontal with speed $u$ has its range equal to $R$ . At what angle should it be thrown with same speed for its range to be half of its initial value.

15^{\circ}

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60^{\circ}

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70^{\circ}

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30^{\circ}

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Solution

The correct option is B $60^{\circ}$
Initial Condition:
$R = \frac{u^{2} sin (2 \times 45^{\circ})}{g} = \frac{u^{2}}{g}$

For range $\frac{R}{2}$

$\frac{u^{2}}{2 g} = \frac{u^{2} sin 2 θ}{g}$

$sin 2 θ = \frac{1}{2}$

$\Rightarrow θ = 15^{\circ}$

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A particle thrown at angle 45∘ with horizontal with speed u has its range equal to R. At what angle should it be thrown with same speed for its range to be half of its initial value.

The correct option is B 60∘Initial Condition: R=u2sin(2×45∘)g=u2g For range R2 u22g=u2sin2θg sin2θ=12 ⇒θ=15∘

A particle thrown at angle $45^{\circ}$ with horizontal with speed $u$ has its range equal to $R$ . At what angle should it be thrown with same speed for its range to be half of its initial value.

The correct option is B $60^{\circ}$
Initial Condition:
$R = \frac{u^{2} sin (2 \times 45^{\circ})}{g} = \frac{u^{2}}{g}$

For range $\frac{R}{2}$

$\frac{u^{2}}{2 g} = \frac{u^{2} sin 2 θ}{g}$

$sin 2 θ = \frac{1}{2}$

$\Rightarrow θ = 15^{\circ}$