A boy standing in a ditch throws a baseball upward toward his father. The ball leaves his hand at ground level, with an initial speed of 15.0 m/s, at an angle of theta = 53.0 degrees from the horizontal. The boy's father reaches up and catches the ball over his head, at a height of 2.0 m above the ground. The father catches the ball on its way down. Calculate how long the ball is in the air. ( g = 9.81 m/s2)
physics - Henry, Monday, October 17, 2011 at 2:39pm
Vo = (15m/ws,53deg.).
Xo = 15cos53 = 9.0mm/s.
Yo = 15sin53 = 12.0m/s.
t(up) = (Yf - Yo) / g,
t(up) = (0 - 12) / -9.81) = 1.22s.
Yf^2 = Yo^2 + 2gh,
h = (Yf^2 - Yo^2) / 2g,
hmax = (0 - (12)^2) / -19.62 = 7.34m.
h = Yo*t + 0.5g*t^2 = (7.34 - 2)m.
0 + 4.9t^2 = 5.34,
t^2 = 1.09,
t(dn) = 1.04s.
T = t(up) + t(dn) = 1.22 + 1.04 = 2.26s
= Time in air.