Posted by **Rodrigo** on Monday, September 17, 2012 at 4:25pm.

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**, Monday, September 17, 2012 at 4:47pm
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

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