posted by Aria on .
A 0.2 kg baseball located at the top of a building which is 100 m above the ground is thrown with an initial velocity 30 m/s at an angle of 30o above the horizontal.
(a) Determine the total energy which the ball will have just after the instant that the ball is thrown.
(b) Determine the maximum height above the ground which the ball will reach during its flight.
a. E = 0.5m*Vo^2 = 0.1*30^2 = 90 Joules.
b. Vo = 30m/s @ 30 Deg.
Xo = hor = 30*cos30 = 26m/s.
Yo = ver = 30*sin30 = 15 m/s.
hmax = ho + (Y^2-Yo^2)/2g
hmax = 100 + (0-225) / -19.6 = 111.5 m.