The maximum range of a projectile is 48 m- When it is thrown at an angle of 15- with the horizontal- its range will be-A- 24 m B- 22 mC- 32 mD- 28 m

The correct option is A 24 mRange of a projectile is R= u^2 sin 2 θg Maximum range is R_max = u^2g Given nbsp; R_max =48=u^2g for θ = 15^∘ R=u^2 sin 30^∘g=u^ ...

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Standard XII

Physics

Range

The maximum r...

Question

The maximum range of a projectile is $48 m$ . When it is thrown at an angle of $15^{\circ}$ with the horizontal, its range will be-

22 m

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24 m

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28 m

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32 m

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Solution

The correct option is B $24 m$
Range of a projectile is $R = \frac{u^{2} sin 2 θ}{g}$
Maximum range is $R_{max} = \frac{u^{2}}{g}$
Given $R_{max} = 48 = \frac{u^{2}}{g}$
for $θ = 15^{\circ}$
$R = \frac{u^{2} sin 30^{\circ}}{g} = \frac{u^{2}}{g} \times \frac{1}{2} = 48 \times \frac{1}{2} = 24 m$

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The maximum range of a projectile is 48 m. When it is thrown at an angle of 15∘ with the horizontal, its range will be-

The correct option is B 24 mRange of a projectile is R=u2 sin 2θg Maximum range is Rmax=u2g Given Rmax=48=u2g for θ=15∘ R=u2 sin 30∘g=u2g×12=48×12=24 m

The maximum range of a projectile is $48 m$ . When it is thrown at an angle of $15^{\circ}$ with the horizontal, its range will be-

The correct option is B $24 m$
Range of a projectile is $R = \frac{u^{2} sin 2 θ}{g}$
Maximum range is $R_{max} = \frac{u^{2}}{g}$
Given $R_{max} = 48 = \frac{u^{2}}{g}$
for $θ = 15^{\circ}$
$R = \frac{u^{2} sin 30^{\circ}}{g} = \frac{u^{2}}{g} \times \frac{1}{2} = 48 \times \frac{1}{2} = 24 m$