This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Let first ball is thrown at t = 0s.
second ball is thrown at t = 2s
u = 40 m/s
g = -10 m/s²

let they meet after t seconds at a height H. Then both the balls will be at H.
for 1st ball, H = ut+(1/2)gt²
for 2nd ball, H = u(t-2) + (1/2)g(t-2)²

ut+ \frac{1}{2}gt^2 =u(t-2)+ \frac{1}{2}g(t-2)^2 \\ \\ut+ \frac{1}{2}gt^2 =ut-2u+ \frac{1}{2}g(t^2+4-4t) \\ \\ut+ \frac{1}{2}gt^2 =ut-2u+ \frac{1}{2}gt^2+2g-2gt \\ \\-2u+2g-2gt=0\\ \\u-g+gt=0\\ \\40-(-10)-10t=0\\ \\10t=50\\ \\t=  \frac{50}{10}=5\ seconds

H = ut+ \frac{1}{2}gt^2 \\ \\H=40*5- \frac{1}{2}*10*5^2=\boxed{75\ m}
Thanks my answer was coming 60m thanks for answering it