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 ?

Step-by-step solution

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 + 1 2 a y t 2 t = 2 u y a y = 2 × 20 10 = 4   s Taking the right-side horizontal as positive x-direction s x = 0 = u x t + 1 2 a x × t 2 0 = u x 4 - 8 2 4 2 u x = 16 tan ⁡ θ = u x u y = 16 20 10 tan ⁡ θ = 10 × 16 20 =   8