A projectile fired at an angle of 15^ ^ ^ achieves a range of 50 m . If the projectile is fired with the same speed at an angle of 45^ ^ ^ , its range will be

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Correct answer: D. 100 m

The horizontal range of a projectile is given by

R = \dfrac{u^2 \sin(2\theta)}{g}

.

For the first case,

\theta = 15^{}^{{}^{\circ}}

and

R = 50 \text{ m}

.

50 = \dfrac{u^2 \sin(2 \times 15^{}^{{}^{\circ}})}{g} = \dfrac{u^2 \sin(30^{}^{{}^{\circ}})}{g}

Since

\sin(30^{}^{{}^{\circ}}) = \dfrac{1}{2}

, we get:

\dfrac{u^2}{2g} = 50 \Rightarrow \dfrac{u^2}{g} = 100 \text{ m}

For the second case, the projectile is fired with the same speed at an angle

\theta' = 45^{}^{{}^{\circ}}

.

R' = \dfrac{u^2 \sin(2 \times 45^{}^{{}^{\circ}})}{g} = \dfrac{u^2 \sin(90^{}^{{}^{\circ}})}{g}

Since

\sin(90^{}^{{}^{\circ}}) = 1

, we get:

R' = \dfrac{u^2}{g} = 100 \text{ m}

A projectile fired at an angle of 15^ ^ ^ achieves a range of 50 m...

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