Physics questions （Kinematics）有answers 但need steps?

1. Use equation of motion: s = ut + (1/2)at^2
For the 1st stone: s1 = (0)t + (1/2)gt^2
For the 2nd stone: s2 = (0)t + (1/2)g(t - 1)^2

Separation between stones y = s1 - s2 = (g/2)[t^2 - (t-1)^2]
i.e. y = (g/2).(2t -1)
Thus, the separation between stones increases with time of fall t. The longer the time, the larger the separation is. The longest time T is when the 1st stone hits the ground. This is give by,

80 = (1/2)g(T)^2
Take g, the acceleration due to gravity, equals to 9.81 m/s^2. Solve for T gives T = 4.039 s

Hence, max separation y(max) = (g/2).(2 x 4.039 -1) m = 34.7 m
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2. Use equation of motion: s = ut + (1/2)at^2
In the 1st second ; s1 = (0).(1) + (1/2)g.(1^2)
i.e. s1 = g/2

At the end of 2 s: s2 = (0)(2) + (1/2)g(2^2)
i.e. s2 = 2g

Thus, distance travelled by the stone in the 2nd second = (2g - g/2) = 3g/2
Ratio of distance travelled in 1st second to that travelled in 2nd second
= (g/2)/(3g/2)
= 1/3 = 1:3

1. the velocity of P is always bigger than Q, So the biggest separartion is that when P reaches the ground. calculate the time of travel of P. Then minus 1, Use this tie to calculate the distance travel by Q. Then 80m minus this distance, The answer comes.
2. s1/ (s2-s1)= Ans, s1= 1/2at1^2 , s2 =1/2at2^2, t1=1s, t2 =2s