Question 6 0 / 7 points A ball is kicked at 10.5 m/s at an angle of 31" to the horizontal. a) How long (time) is the ball in the air? (4 marks) b) Find the horizontal displacement (range) of the ball. (2 marks) c) How far will the ball have travelled horizontally, when it reaches its maximum height? (1 mark) - No text entered - This question has not been graded. The correct answer is not displayed for Written Response type questions. Hide question 6 feedback a) Take up as positive direction Viy = (10.5 m/s)sin310 = 5.4 m/s a = -9.8m/82 Ady = 0 (back to ground level - - > so, no vertical displacement) At =? Ad = vAt+ za(At)? 0 = ((10.5 m/s)sin31')At + 2 (-9.8 m /s?) (At)? 0 = (5.4 m/s)At - 4.9m/s2(At)2 At =1.1$ bj Ads = VzAt Adr = ((10.5 m/s)cos310) (1.1 s)using the unrounded value Adr = 9.9 m c) Half the distance of answer b), which means that at its maximum height, it will have travelled 5 m