A boy throws a ball at an angle θ with the vertical. If the vertical component of the initial velocity is 20m s ^ - 1 and the wind imparts a horizontal acceleration of 8m s ^ - 2 to the left, the angle at which the ball

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Correct answer: 8

Taking the upward vertical as positive y-direction

{u}_{y} = 20m{s}^{- 1}

{a}_{y} = - 10m{s}^{- 2}

{s}_{y} = {u}_{y}t + \dfrac{1}{2}{a}_{y}{t}^{2}

t = \dfrac{2{u}_{y}}{{a}_{y}} = \dfrac{2 \times 20}{10} = 4s

Taking the right-side horizontal as positive x-direction

{s}_{x} = 0 = {u}_{x}t + \dfrac{1}{2}{a}_{x} \times {t}^{2}

0 = {u}_{x}\left(4\right) - \dfrac{8}{2}{\left(4\right)}^{2}

{u}_{x} = 16

tan \theta = \dfrac{{u}_{x}}{{u}_{y}} = \dfrac{16}{20}

10tan \theta = 10 \times \dfrac{16}{20} = 8

A boy throws a ball at an angle θ with the vertical. If the...

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