Posted by tank on Monday, June 6, 2011 at 11:57pm.
A ball is thrown upward from a platform 4.5 m high with a speed of 17 m/s at an angle of 41 o from the horizontal. What is the magnitude of its velocity when it hits the ground?

physics  Henry, Tuesday, June 7, 2011 at 6:16pm
Vo = 17m/s @ 41deg.
Vo(h) = 17cos41 = 12.8m/s.
Vo(v) = 17sin41 = 11.15m/s.
Vf^2 = Vo^2 + 2gd = 0.
(11.15)^2 + (19.6)d = 0,
124.32  19.6d = 0,
19.6d = 124.32,
d2 = 6.34m = distance above the platform.
d = d1 + d2,
d = 4.5 + 6.34 = 10.84m = Distance above ground.
Vf^2 = Vo^2 + 2gd,
Vf^2 = 0^2 + 2*(9.8)*10.84 = 212.52,
Vf = 14.6m/s = Final velocity.
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