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- 天同Lv 75 years agoFavorite Answer
(a)(i) Acceleration from A to B = 20/4 m/s^2 = 5 m/s^2

Deceleration from C to D = 20/5 m/s^2 = 4 m/s^2

(ii) Total distance travelled = area of trapezium = (6 + 15) x 20/2 m = 210 m

(iii) Average speed = 210/15 m/s = 14 m/s

(b)(i) Use equation v = u + at

0 = 25 + (-10)t [take g = -10 m/s^2]

i.e. t = 2.5 s

(ii) Height from the starting point that the ball reached = 25^2/2g = 31.25 m

Hence, max height from ground reached by the ball = (31.25 + 50) m = 81.25 m

(iii) Height from starting point that the ball should reach

= (95 - 50) m = 45 m

Use equation: v^2 = u^2 + 2as

0 = u^2 + 2(-10).(45)

i.e. u = 30 m/s

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