Posted by amy on Tuesday, September 18, 2012 at 12:47pm.
evidently the horizontal component of the ball's velocity is 18.0m/s toward the rear of the train.
18.0 = v*cos68.0°
v = 48m/s
The vertical component is thus 48*sin68.0° = 44.5 m/s
the height above the train is thus
h = 44.5t - 4.9t^2
= t(44.5 - 4.9t)
h is max at t=4.54
h(4.54) = 101m
can yu help me thanks in advanec.
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