Basic kinematics

A cannon shoots at 400m/s towards a target 400m high and 15000m away from it, what angles must the cannon be a
t?

this is a kinematics question i have for homework and just dont know where to turn to. i tried with the equations but seem to always have something missing

 

i have tried everything i can using

v(t)=vo+at
x(t)=Xo+Vot+0.5at^2
v^2-Vo^2=2a(delta)X
or other variations of these equations

every time i come up with 1 equation with 2 missing numbers and the i get stuck, for example, using the 1st and saying that at max height Vy0=0, i get stuck because i dont have sin(theta) or T
using the 2nd and saying that Yf=400 i still dont have sin(theta) or T
using the 2nd and saying Xf=15000 dont have cos(thetha) or t
ising the 3rd dont have V unless i say Vy at max height is 0 but then dont have delta Y

 

still dont know how to do it, what am i supposed to do, i found that ymax height=8163 x sin(theta), dont see how that helps me,

please help, i have another 4 questions of this sorty and cant continue my homework till i understand it, my next class is on monday and need to finish the work before it

 

i did that, i get , for x- t=3.75/cos(theta)
then i must put that same t into my y equation, but how do i get from cos(theta) to sin(theta)??

do you understand why i cant solve this or am i just being an idiot

 

for x
15000=400xcos(th)xt
t=37.5/cos(th)
cos(th)=37.5/t

for y
400=400xsin(th)xt-4.9t^2
t=81.633sin(th)
sin(th)=0.01225t

now from this how do i find theta??

 

kurdt, the bold doesnt really matter,


my problem is that if i substitute the t i found from the X, i get an equation with sin(theta) and cos(theta) which i think i can turn into tan(???) but i also get cos^2 from the t^2 and then dont know how to bring them to a result for theta

400=400sin(th)x37.5/cos(th)-4.9x(37.5/cos(th))^2