Question

Ala A ball is launched from point A at angle 30° as shown in Fig. Ala. Determine the maximum and minimum speed vA it can have so that it lands in the container [10 marks] 30° 1 m 0.25 n 2.5 m 4 m

Answer 1

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theta = 30 degrees g = 9.8 m/s^2 If Range = 2.5 m then v_A cos theta(t) = 2.5 -0.75 = v_A sin theta t - 1/2 gt^2 -0.75 = v_A sin theta(2.5/vA cos theta) - 1/2 gt^2 1/2 gt^2 = 2.5 tan theta + 0.75 t = 0.67 sec For Range = 4m v_A^' cos theta(t^') = 4 -0.75 = v_A^' sin theta t^' - 1/2 g(t^')^2 1/2 g(t^')^2 = 4 tan theta + 0.75 t^' = 0.7902 sec For Range = 2.5 m (v_A cos theta)t = 2.5 v_A = 2.5/0.67 cos 30 degrees v_A = 4.31 m/s For Range = 4m (v_A^' cos theta)t^' = 4 v_A^' = 5.845 m/s maximum speed = 5.845 m/s minimum speed = 4.31 m/s