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Step2247 [10]
3 years ago
12

A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed o

f the horse is 10.0 m/s, and the distance from the limb to the level of the saddle is 3.00 m. (a) What must be the horizontal distance between the saddle and limb when the ranch hand makes his move? (b) For what time interval is he in the air?
Physics
1 answer:
Natali5045456 [20]3 years ago
3 0
The time it will take him to fall can be found from:-3m = -(g*t^2)/2
Find that time, it's the time the horse will travel horizontal while the cowboy is falling.So the horizontal distance away is  10 m/s * t
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Answer:

Calculating Coefficient of friction is 0.229.

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Explanation:

We know that speed expression is as \mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}.

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Cancel "m" common term in both L.H.S and R.H.S

\text { Equation becomes, } \mu \times g=a

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\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })

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