Suppose the departure angle in Fig. 3.26 is 42.0° and the distance d is 3.00 m. Where will the dart and monkey meet if the initial speed of the dart is (a) 12.0 m/s? (b) 8.0 m/s? (c) What will happen if the initial speed of the dart is 4.0 m/s? Sketch the trajectory in each case.

Solution 20E (a) When the initial speed of the dart is = 12.0 m/s 0 Its x-component is = 12.0 × cos 42 v x 8.92 m/s The y-component of velocity is v = 12.y m/s × sin 42 0 v y 8.02 m/s The horizontal distance is = 3.00 m Let the time to reach his distance is t. Therefore, v t x 3.00 m t = 3.00/8.92 m/s t = 0.336 s Now, the height moved by the dart when it hits the monkey is h = v t gt 1 2 1 y 2 1 2 h 1 8.02 × 0.336 m 2 × 9.8 × (0.336) m h 1 2.14 m So, the dart and the monkey will meet at a height of 2.14 m when the speed of the dart is 12.0 m/s. (b) When the speed of the dart is 8.0 m/s. 0 Its x-component v = 8x0 × cos 42 v x 5.95 m/s The y-component of velocity is v = 8.0y× sin 42 0 v = 5.35 m/s y Let t be the time for the dart to hit the monkey, v t = 3.00 m x t...