Question

A shot-putter 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?

Answer 1

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.