first vertical problem
45 = 4.9 t^2
t = 3.03 seconds to fall
distance ball is from building = 3.03*5
= 15.15
distance man must run = 15.15 - 6 = 9.15
time man has = 3.03-.3 = 2.73 seconds
9.15 = .5 a (2.73)^2
solve for a
A ball is launched horizontally at 5 m/s from the top of a 45 m tall building. A man stands at rest 6 m from the base of the building. If there is a 0.3 s reaction time for the man to start moving once he sees the ball launched, find the magnitude of the man's acceleration needed to catch the ball?
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