maths

1)
a = 10 m/s^2
s(2 sec) = 0
s(3 sec) = 25m
s = s0 + ut + (1/2)*at^2
s(2) = s0 + 2u + 5*4
0 = s0 + 2u + 20
s(3) = s0 + 3u + 5*9
25 = s0 + 3u + 45
0 = s0 + 3u + 20
u = 0
s0 = -20
The body initial displacement relative to O is 20m to O.
2(a)
v(t) = u + at
v(t) = 25 + (-10)t
v(t) = 25 - 10t
2(b)
s(t) = ut + (1/2)*at^2
s(t) = 25t +1/2(-10)t^2
s(t) = 25t - 5t^2
2(c)
At maximum height, v(t) = 0 where t is bigger than zero
25 - 10t = 0
10 t = 25
t = 2.5 sec
2(d)
maximum height = s(2.5) = 25(2.5) -5 (2.5)^2
= 62.5 - 31.25
= 31.25 m
2(e)
the time taken to return to the point of projection = t
s(t) = 0
25t - 5t^2 = 0
25 t = 5 t^2
t = 5 or t = 0 (rejected)
the time taken to return to the point of projection is 5 sec
I hope this can help with your understanding. =)

this is about physics - linear motion