Question

Lily throws a ball at an angle of 53.13 degrees above the horizontal from a building that is 10m high. The ball has a speed of 12m/s, .6 seconds after she throws it. What was the initial speed of the ball? This is a vector problem and uses variables such as vx, vy, vyi, delta y and delta x

Answer 1

Vinitial = Vi
vx = x velocity = constant = Vi cos 53.13

vyi = Vi sin 53.13


at t = .6
vx =vx still
vy = vyi - g t
so
V at.6=12 = sqrt {vx^2+ (Vyi-9.81*.6)^2]