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

Question

A boy throws a ball at an angle θ with the vertical. If the vertical component of the initial velocity is 20 m s - 1 and the wind imparts a horizontal acceleration of 8 m s - 2 to the left, the angle at which the ball must be thrown so that the ball returns to the boy's hand is θ . What is the value of 10 tan ⁡ θ ?

TestHub · Accepted Answer

Taking the upward vertical as positive y-direction

$u_{y} = 20 m s^{- 1}$

$a_{y} = - 10 m s^{- 2}$

$s_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2}$

$t = \frac{2 u_{y}}{a_{y}} = \frac{2 \times 20}{10} = 4 s$

Taking the right-side horizontal as positive x-direction

$s_{x} = 0 = u_{x} t + \frac{1}{2} a_{x} \times t^{2}$

$0 = u_{x} (4) - \frac{8}{2} {(4)}^{2}$

$u_{x} = 16$

$\tan θ = \frac{u_{x}}{u_{y}} = \frac{16}{20}$

$10 \tan θ = 10 \times \frac{16}{20} = 8$

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