Asked by tab
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)
Answers
Answered by
Henry
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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.