Posted by Chan on Wednesday, September 30, 2009 at 1:27am.
A. Multiply the horizontal component of the initial velocity by the time that the ball is in the air.
7.80 x cos 21 x 3.00 = ?
B. Solve the following equation for vertical height above the ground, for the unknown height H.
y = 0 = H -7.80 sin 21 t - (g/2) t^2
using t = 3.0 seconds. g is the acceleration of gravity, 9.8 m/s^2.
C. Use a different vertical height equation, with t as the unknown and the vertical height above the ground being
y = H -10
= H -7.80 sin 21 t - (g/2) t^2
Note that the H cancels out, leaving
-10 = -7.80 sin 21 t - (g/2) t^2
There will be two roots. Take the one that is positive
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