A golf ball rolls off a horizontal cliff with an initial speed of 11.4 m/s. The ball falls a vertical distance of 17.7 m into a
1 answer:
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The value of the force, F₀, at equilibrium is equal to the horizontal
component of the tension in string 2.
Response:
- The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>
Given:
The weight of the rod = The sum of the vertical forces in the strings
Therefore;
M·g = T₂·cos(37°) + T₁
The weight of the rod is at the middle.
Taking moment about point (2) gives;
M·g × L = T₁ × 2·L
Therefore;
Which gives;
F₀ = T₂·sin(37°)
Which gives;
- F₀ ≈ <u>0.377·M·g</u>
<u />
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Answer:
The height of the cliff is 90.60 meters.
Explanation:
It is given that,
Initial horizontal speed of the stone, u = 10 m/s
Initial vertical speed of the stone, u' = 0 (as there is no motion in vertical direction)
The time taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s
Let h is the height of the cliff. Using the second equation of motion in vertical direction to find it. It is given by :
h = 90.60 meters
So, the height of the cliff is 90.60 meters. Hence, this is the required solution.