re: HSC Physics Marathon Archive
I'm not too sure about my answer.
Let y=0 be ground level
y=1/2 a(y)t^2 + u(y) +100
Uy=0, as it is released horizontally---> So y=1/2 a(y)t^2 + 100 (1)
x= U(x)t, Ux=10---> x=10t (2)
tan45=y/x, so y=x
Equating (1) and (2), we get a quadratic equation...