Question

A ball is thrown at angle theeta anf another ball is thrown at angle (90-theeta) with the horizontal direction ftom the same point each with speed of 40m/s . The second ball reaches 50m higher than the first ball. Finf their individual hights.g=9.8m/s^2

Answer 1

As  time is not given we have to use formula
v^2 = u^2 -2aS
u =39.2Sinθ,v=0
so
1536.64Sin^2(θ) = 19.6S-----(1)

for second ball
u=39.2Sin(90-θ),v=0
1536.64Sin^2(90-θ) = 19.6(S+50)-----(2)

adding the both equation 

1536.64{Sin^2(θ) +Sin^2(90-θ)} = 39.2S + 980
as   Sin^2(90-θ) = Cos^2(θ)
and Sin^2(θ) + Cos^2(θ) = 1

1536.64 - 980 = 39.2S
S = 14.2 m
S+50 = 64.2 m

Height of rst ball is 14.2m and height of second ball is 64.2m

i hope it will help you
regards