At a particular instant a hot air balloon is 100m in the air and decending at a constant speed of 2m/s at this exact instant a girl throws a ball horizontally relative to herself with an initial speed of 20m/s when she lands where will she find the ball?

Explanation:

Consider the vertical motion of ball

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 2 m/s

        Acceleration, a = 9.81 m/s²  

        Displacement, s = 100 m      

     Substituting

                      s = ut + 0.5 at²

                      100 = 2 x t + 0.5 x 9.81 xt²

                      4.905t²  + 2t - 100 = 0

                     t = 4.32 s    or   t = -4.72 s

                    After 4.32 seconds the ball reaches ground.

Now we need to find horizontal distance traveled by ball in 4.32 seconds.

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 20 m/s

        Acceleration, a = 0 m/s²  

        Time, t = 4.32 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 20 x 4.32 + 0.5 x 0 x 4.32²

                      s = 86.4 m

She will find the ball at a horizontal distance of 86.4 m from landed location

i don't know that i agree with that but what's the question ?

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle