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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