For projectile motion at 45°, vertical component of initial velocity:
v0y=v0sin45°=2v0
Maximum height occurs at t=2 s when vy=0:
0=v0y−gt
⇒v0y=10×2=20 m/s
Thus 2v0=20, so v0=202 m/s
Height at landing time t=3 s:
H=v0yt−21gt2=20×3−21×10×9
=60−45=15 m