Posted by JLF on .
A shotputter releases the 16lbs shot at an angle of 41 degrees/ Of the shot travels a horizontal distance of 34m to reach its peak at midflight, how high did the shot go and what is the displacement of the resulting vector at midflight?

Math 
Steve,
y = x tanθ  g/(2 (v cosθ)^2) x^2
= .86x  8.6/v^2 x^2
y reaches its max when x = 5/v^2
v^2/20 = 34
v = 26
y = .86x  .0126x^2
y(34) = .86(34)  .0126 (34^2) = 14.67
Now just figure the displacement and you're done.