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Question
In the following figure shows a 11.7 ft wide ditch with the approach roads at an angle of 15° with the horizontal. With what minimum speed should a motorbike be moving on the road so that it safely crosses the ditch?
Assume that the length of the bike is 5 ft, and it leaves the road when the front part runs out of the approach road.
Solution
Given:
Width of the ditch = 11.7 ft
Length of the bike = 5 ft
The approach road makes an angle of 15˚ (α) with the horizontal.
Total horizontal range that should be covered by the biker to cross the ditch safely, R = 11.7 + 5 = 16.7 ft
Acceleration due to gravity, a = g = 9.8 m/s = 32.2 ft/s2
We know that the horizontal range is given by
\[R = \frac{u^2 \sin2\alpha}{g}\]
By putting respective values, we get:
\[u^2 = \frac{Rg}{\sin2\alpha} = \frac{16 . 7 \times 32 . 2}{\sin30^\circ}\]
\[ \Rightarrow u \approx 32 ft/s\]
Therefore, the minimum speed with which the motorbike should be moving is 32 ft/s.
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