Let v be the unknown speed. We will deduce distance in terms of v, make it equal to 1540 and solve for v.

The position of a projectile at time t in a vector form may be written as r=<v*cos(37*pi/180)*t , v*sin(37*pi/180)*t - (9.8*t^2)/2>

Making the second component equal to zero, we find t>0 at which projectile reaches the ground. v*sin(37*pi/180)*t - 4.9*t^2 = 0, so t=v*sin(37*pi/180)/4.9.

Substituting this into firs component of position and making it equal to 1540 we get equation v*cos(37*pi/180)*(v*sin(37*pi/180)/4.9)=1540 or

(cos(37*pi/180)*sin(37*pi/180)/4.9)*v^2 = 1540

solving for v we get

v=1540*4.9/(cos(37*pi/180)*sin(37*pi/180)) ~ 125.30 m/s