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/s2 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)

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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