A stone is thrown from the top of a building TT toward another building 50m away. The initial velocity of the stone is 20m/s, at 400 above horizontal. How far above or below its original level will the ball strike the opposite wall?
"v_{x0}=20\\cdot\\cos40\u00b0=15.32(m\/s)"
"v_{y0}=20\\cdot\\sin40\u00b0=12.86(m\/s)"
The stone flight time
"t=50\/15.32=3.264(s)"
"y=v_{y0}t-gt^2\/2=12.86\\cdot 3.264-9.8\\cdot3.264^2\/2=-10.23(m)"
"10.23\\ (m)" below its original level. Answer
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