How do I calculate the maximum altitude in this projectile motion problem?

The formula for horizontal range is x = Vo x Costhetha x Time

Okay, lets answer your question

Y of a projectile motion is at the instant where y' = 0
and we know that  y' = -gt + Vo.SinTheta
Since Y is maximum ( the maximum altitude) at instant where y' = 0
i will put 0  in the equation y' = -gt + Vo.SinTheta instead of y'
==>
-gt + Vo.SinTheta =0
==>
t = VoSintheta / g = (20 x sin40  )/ 10= 1.28557 seconds
=====>
Therefore
y(max) = -0.5 x g  x t^2 + (Vo x  Sinthetha x T)
= -0.5 x 10 x 1.28557^2 + (20 x 0.64278 x 1.28557)
= -8.263451125 + 16.52696935
= 8.263518226 m

Therefore y maximum is 8.263 meters:D
hope i helped you:D