Question

A ball is projected with a speed of 10 m/s. The two angles of projection for which the range is 5m are ? (g=10 m/s^2)

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Angle\:of\:projection=15°}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a ball is projected with a speed of 10 m/s.

• We have to find two angles of projection for which the range is 5m .

$\underline \bold{Given : }\\ \implies Initial \: speed(u) = 10 \: ms \\ \\ \implies Range = 5 \: m \\ \\ \implies Gravity(g) = 10 \: m {s}^{2} \\ \\ \underline \bold{To \: Find : } \\ \implies Angle \: of \: projection( \theta) = ?$

• According to given question :

$\bold{Using \: formula \: of \: range : } \\ \implies R = \frac{ {u}^{2} \: sin \: 2 \theta }{g} \\ \\ \implies 5 = \frac{10 \times 10 \times sin \: 2\theta}{10} \\ \\ \implies \frac{\cancel{50}}{\cancel{100}} = sin \: 2 \theta \\ \\ \implies sin \: 2 \theta = \frac{1}{2} \\ \\ \implies sin \: 2 \theta = sin \: 30 \degree \\ \\ \implies 2 \theta = 30 \degree \\ \\ \implies \bold{\theta = 15 \degree}$

Answer 2

$\large \bold{ \underline{ \underline{ \sf \: Answer : \: \: \: }}}$

$\to \sf Angle \: of \: projection = 15 \degree$

$\large \bold{ \underline{ \underline{ \sf \: Explaination : \: \: \: }}}$

Given ,

Speed ( u ) = 10 m/s

Range ( R ) = 5 m

Acceleration due to gravity ( g ) = 10 m/s²

We know that ,

$\large \bold{ \fbox{ \fbox{ \sf Range = \frac{ {(u)}^{2} \: \times \: sin2 \theta }{g} }}}$

$\to \sf 5 = \frac{ {(10)}^{2} \times sin2 \theta }{10} \\ \\ \to \sf 5 = \frac{100 \times sin2 \theta }{10} \\ \\ \to \sf sin2 \theta = \frac{5}{10} \\ \\ \to \sf sin2 \theta = sin \: 30 \degree \\ \\ \to \sf 2 \theta = 30 \degree \\ \\ \to \sf \theta = 15 \degree$

$\therefore$ The required value of angle of projectionis is 15°

_______ Tera Bhai ❤