A projectile is thrown at an angle of 30 degree with the horizontal so that its range is 200m.Aother projectile is thrown at an angle 60 degree with the vertical .Calculte its range.

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Solution

$R = \frac{u^{2} \sin^{2} 30}{g} R' = \frac{u^{2} \sin^{2} 60}{g} \frac{R'}{R} = \frac{\frac{u^{2} s i n^{2} 60}{g}}{\frac{u^{2} s i n^{2} 30}{g}} = \frac{3 / 4}{1 / 4} = 3 R' = 3 \times 200 = 600 m$

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A projectile is thrown at an angle of 30 degree with the horizontal so that its range is 200m.Aother projectile is thrown at an angle 60 degree with the vertical .Calculte its range.

R = u2sin230gR' = u2sin260gR'R=u2sin260gu2sin230g = 3/41/4 = 3R' = 3×200 = 600 m

$R = \frac{u^{2} \sin^{2} 30}{g} R' = \frac{u^{2} \sin^{2} 60}{g} \frac{R'}{R} = \frac{\frac{u^{2} s i n^{2} 60}{g}}{\frac{u^{2} s i n^{2} 30}{g}} = \frac{3 / 4}{1 / 4} = 3 R' = 3 \times 200 = 600 m$