[Mikez]

Person A standing on the roof throws a book up vertically at 10m/s. At the same time, another person below that person 12 m throws a ball at 25 m/s. 

when will the 2 objects meet and how far above the roof do they meet?

-can anyone help me answer this question?

[Borek]

Accelerated motion (or rather decelerated motion) in both case with the same acceleration (decelaration). Different v₀ and x₀. Two functions of time, compare them and solve for t.

[Mikez]

I tried doing that....   since D=ST,     10 m/s*t=25 m/s*t  ....but I couldn't incorporate the fact that there was a 12 m difference.

[Borek]

To say the truth I have no idea what is nomenclature used in English when doing such questions

Write equation linking distance with time, acceleration and starting veocity. And name symbols you use to help me understand

[Donaldson Tan]

when the 2 objects meet, they share the same displacement (s), so
s_book = s_ball

the equation for displacement s for constant acceleration is s = u.t + 0.5*a.t²
where u is initial velocity, a is acceleration and t is time

using the ground as the reference point,
s_book = 10*t + (1/2)(-g)t² + 12 = 10*t - (1/2)(9.81)t² + 12
s_ball = 25*t + (1/2)(-g)t² + 0 = 25*t - (1/2)(9.81)t²

let s_book = s_ball
10*t - (1/2)(9.81)t² + 12 = 25*t - (1/2)(9.81)t²
10*t + 12 = 25*t
15*t = 12
t = 12/15 = 0.8 second

Both objects meet at time 0.8s

displacement of both objects = s_ball( t=0.8 ) = 25*( 0.8 ) - (1/2)(9.81)( 0.8 )² = 16.86m

Height above roof = 16.86 - 12 = 4.86m

[agadou]

In a referenceframe accelerating downward by g fix another referenceframe to the body thrown up by 10 m/s. In that frame the speed of the body thrown up from below 12m is 15 m/s and has to travel 12  meter to meet the other. t=s/v, t=12/15=0,8s