An archer shoots an arrow into the air with an initial velocity of 75 feet per second. The initial height of the shot is 5 feet. The arrow strikes the ground precisely 200 feet from the archer. At what angle did the archer shoot the arrow?

This problem has a solution only if we assume the vertical component of the initial velocity is 75 ft/s. The time the arrow is in the air is then the solution to
  0 = -16t^2 + 75t + 5    (the equation for the vertical height with height set to zero)
By any suitable method, the solution of interest is t = (75+√5945)/32 ≈ 4.75324 seconds.

Then the horizontal component of the velocity is
  (200 ft)/(4.75324 s) ≈ 42.0765 ft/s

The angle is the arctangent of the ratio of the vertical to horizontal velocities
  angle = arctan(75/42.0765) ≈ 60.71° _____
If the total velocity is restricted to 75 ft/s, the maximum range is 180.712 ft at a launch angle of 44.21°. The velocity must be increased to about 79.01 ft/s in order to get a range of 200 ft. We conclude, therefore, that the simpler problem with a given vertical velocity is the problem intended.