Urgent physics question answer ASAP!!!?

2018-06-27 1:04 pm
Please help me. I have tried solving for knowns myself, asking others, but none of them were able to help me crack this problem. A worked out solution to this problem will mean the world to me. Thanks!
更新1:

A stone at the end of a sling is whirled in a vertical circle of radius r = 1.14m. The center of the sling is at a height h = 1.75m above the ground. The stone is released at point A when the sling is inclined at 30.0 degrees above horizontal. If the centripetal acceleration of the stone immediately before release is 36.7m/s^2, at what horizontal distance in meters does the stone land as measured from the point directly below the center of the circle?

回答 (4)

2018-06-27 2:51 pm
✔ 最佳答案
Refer to the diagram below.

Centripetal acceleration = vₒ² / r
vₒ² / 1.14 = 36.7
vₒ = √(36.7 × 1.14) m/s = 6.47 m/s

Consider the vertical motion (uniform acceleration motion):
Take g = 9.81 m/s² and take all downward quantities as positive.
Initial speed, vₒ(y) = -6.47 × cos30° m/s
Acceleration, a(y) = 9.81 m/s²
Displacement, s(y) = (1.75 + 1.14 × sin30°) m

s(y) = vₒ(y) t + (1/2) a(y) t²
1.75 + 1.14 × sin30° = (-6.47 × cos30°) × t + (1/2) × 9.81 × t²
2.32 = -5.6t + 4.905t²
4.905t² - 5.6t - 2.32 = 0
t = 1.465 s or t = -0.323 s

Consider the horizontal motion (uniform velocity motion):
velocity, vₒ(x) = 6.47 × sin30° m/s

Horizontal distance measured from the point directly below the center of the circle when the stone lands
= vₒ(x) t - 1.14 cos30°
= [(6.47 × sin30°) × 1.465 - 1.14 cos30°] m
= 3.75 m
2018-06-28 11:42 am
depends
2018-06-27 1:08 pm
Physics problems can be very difficult, especially when you aren't given the problem.
2018-06-27 1:17 pm
Thats the problem I am given. Please if you can show me a solution to the problem or simply tell me what I need to do to reach the solution because I feel hopelessly confused at this point


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