Question

The horizontal range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 75°, its range will be​

Answer 1

Explanation:

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Answer 2

Answer:

Method :1

$range \: (r) = 50 \: m \\ angle \: of \: projection \: ( \alpha ) = {15}^{0} \\ r = \frac{ {u}^{2} \sin(2 \alpha ) }{g} \\ 50 = \frac{ {u}^{2} \sin( {30}^{0} ) }{g} \\ {u}^{2} = 100g \\ \\ angle \: of \: projection = {75}^{0} \\ r = \frac{100g \sin( {150}^{0} ) }{g} \\ r = 100 \times \frac{1}{2 } \\ r = 50m$

Method :2

$the \: ranges \: of \: the \: projectile \\ \: motion \: having \: same \: initial \\ \: speeds \: but \: different \: angles \\ \: of \: projection (whose \: sum \: is \: {90}^{0} ) \\ is \: equal \\r = \frac{ {u}^{2} \sin(2 \alpha ) }{g} = 50 \: m \\ \\ r = \frac{ {u}^{2} \sin(2(90 - \alpha )) }{g} \\ r = \frac{ {u}^{2} \sin(180 - 2\alpha ) }{g} \\ r = \frac{ {u}^{2} \sin(2 \alpha ) }{g} = 50 \: m$