Posted by Rodrigo on .
A ball thrown into the air lands on the same horizontal level, 31 m away, and 2.65 s later. Find the magnitude of the initial velocity.

Physics 
Steve,
if it rose and fell for 1.325s, then it rose and fell 4.9*1.325^2 = 8.6m.
Now, for a given angle θ and initial velocity v,
range R = v^2 sin(2θ)/g
max height H = v^2 sin^2(θ)/g
R/H = 4cotθ
So, here, 31/8.6 = 4cotθ
cotθ = 0.90
θ = 48°
using the range R,
31 = v^2 sin96°/g
v^2 = 31g/sin96° = 305.5
v = 17.48