A baseball player hits a homerun that rises to a maximum height and just clears a fence that is 17.5 meters high on it’s way back down. The fence is located 116 meters from home plate (where the ball was hit). The ball was hit at an angle of 28.0 degrees above the horizontal from a height of 0.850 meters above the ground.
How long does it take the baseball to reach the fence?
Kinematic question with projectiles?? Please help?
2017-01-20 3:37 pm
回答 (2)
2017-01-20 4:16 pm
✔ 最佳答案
Take g = 9.8 m/s²u : Initial velocity of the baseball
t : Time take for the baseball to reach the fence
The horizontal motion is a uniform velocity motion :
s(x) = v(x) t
116 = (u cos28°) t
u = 116 / (t cos28°) …… [1]
The vertical motion is a uniform acceleration motion :
s(y) = u(y)t - (1/2)gt²
(17.5 - 0.850) = (u sin28°)t - (1/2)(9.8)t²
16.65 = (u sin28°)t - 4.9t² …… [2]
Substitute [1] into [2] :
16.65 = [116 sin28° / (t cos28°)]t - 4.9t²
16.65 = (116 tan28°) - 4.9t²
4.9t² = (116 tan28°) - 16.65
t = √{[116 tan28° - 16.65] / 4.9}
t = 3.0 s
2017-01-20 9:57 pm
sin 28° = 0.4695
cos 28° = 0.8829
17.5 = 0.85+ Vo*sin 28*t-4.903t^2
116 = Vo*cos 28*t
Vo = 116/(0.8829*t)
17.5 = 0.85+116/(0.8829*t)*t-4.903t^2
131.39+0.85-17.5 = 4.903t^2
t = √(131.39+0.85-17.5)/4.903 = 4.838 sec
cos 28° = 0.8829
17.5 = 0.85+ Vo*sin 28*t-4.903t^2
116 = Vo*cos 28*t
Vo = 116/(0.8829*t)
17.5 = 0.85+116/(0.8829*t)*t-4.903t^2
131.39+0.85-17.5 = 4.903t^2
t = √(131.39+0.85-17.5)/4.903 = 4.838 sec
收錄日期: 2021-05-01 13:08:38
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170120073719AAjLAY8