There is a long smooth slope and we release balls from the top of it every 1.5 s

Question

There is a long smooth slope and we release balls from the top of it every 1.5 s. Suppose the acceleration of the ball is 4.0 m/s^2 and every ball is numbered in the releasing order.

a1) 9.5 s after the first ball is release, what is the distance between ball No. 1 and ball No. 6? (m)

a2) What is the distance between ball No. 2 and ball No. 7? (m)

b1) When ball No. 5 reached the speed of 15.0 m/s, what is the distance between ball No. 1 and ball No. 5? (m)

b2) What is the distance between ball No. 2 and ball No. 5? (m)


BTW, a1 answer is not 112.5 or 223.1
a2 answer is not 72 or 166.87
B1 answer is not 40.5 or 180
b2 answer is not 133.125

Explanation / Answer

a1) s1 = 0.5at^2= 0.5x4x9.5^2 = 180.5 m s6 = 0.5x4x(9.5-5x1.5)^2 = 8 m diff= 180.5-8 = 172.5 m a2) s2 = 2x(9.5-1.5)^2 = 128 s7 = 2x(9.5-6x1.5)^2 = .5 diff= 127.5 m b1) v = at 15 = 4t t = 3.75 sec s5 = 2t^2 = 2x3.75^2 =28.125 m ball 1 time = 6+3.75 = 9.75 sec s1 = 2x9.75^2 = 190.125 m = diff = 162 m b2) time for ball 2 when ball 5 has reached 15 m/s = 3*1.5 + 3.75 = 8.25 s2 = 2x8.25^2 = 136.125 m s5 = 28.125 diff = 108 m

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