Question

17. A ball is projected at 45° and its horizontal range in 10 m. velocity of projection is A) 7/3m ms1 C 9.6 ms1 B) 15 ms 1 D), 773m ms1​

Answer 1

Answer:

velocity of projection is about 10

Answer 2

$\sf Given \\\sf\longrightarrow\thetaθ = 45° \\ \sf⟶ R = 10m \\ \sf where \\\sf\longrightarrow\thetaθ \: is \: the \: angle \: of \: projection.\\\sf\longrightarrow R \: is \: range. \\ \sf To \: find - \\\sf Maximum \: height \: \longrightarrow H \:\\\sf Formula used -\\\sf{R = \frac{ {u}^{2} \sin2 \theta }{g} } \\\sf{H = \frac{ {u}^{2} \sin^{2} \theta }{2g} } \\\sf where \\\sf \longrightarrow R \: is \: range.\\\sf\longrightarrow u \: is \: initial\: velocity. \\\sf\longrightarrow g \: is \: acceleration \: due \: to \: gravity. \\\sf\longrightarrow H \: is \: maximum \: height.\\\sf\longrightarrow\thetaθ \: is \: the \: angle \: of \: projection. \\\sf Solution \\\sf Substituting \: the \: value \: in\\\sf{R = \frac{{u}^{2} \sin2 \theta }{g}}\\\implies\sf{10 = \frac{{u}^{2} \sin2(45)}{10}} \\\implies\sf{10 = \frac{ {u}^{2} }{10}}\\\implies\sf{{u}^{2} = 10 \times 10} \\\implies \sf{u = 10 m/s } \\\sf\longrightarrow R = 10 \: m \\\sf\longrightarrow u = 10 \: m/s\\\sf\longrightarrow \thetaθ = 45°\\\sf{Substituting \: the \: value \: in}\\\sf{H = \frac{ {u}^{2} \sin^{2} \theta }{2g}}\\\implies \sf{H = \frac{ {10}^{2} \times \sin^{2} 45}{2 \times 10}}\\\implies\sf{\frac{100}{20} } \\\implies\sf{H = 5m }$