A $11.7ft$ 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 \[5{\text{ }}ft\], and it leaves the road when the front part runs out of the approach road.

Answer
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Hint: In order to answer the question we will assume the bike is a point object, and the ditch is \[2.5{\text{ }} + {\text{ }}2.5 = 0.5\] feet wider, or \[16.7\]feet deep, so it leaves the road as the front portion runs out the approach road. Then, using the definition of a projectile, we will calculate the minimum speed with a projection angle of 15 degrees and a length of 16.7 feet.
Complete step by step answer:
Given, Width of the ditch= $11.7ft$ and Length of the bike= $5ft$. The approach road makes an angle of ${15^ \circ }\left( \alpha \right)$ with the horizontal. The biker's total horizontal width that must be protected in order to safely cross the ditch,
$R = 11.7 + 5 = 16.7\,ft$
Acceleration due to gravity,
$a = g = 9.8\,m/s \\
= \,32.2\,ft/{s^2} \\ $
The horizontal range, as we know, is determined by
$R = \dfrac{{{u^2}\sin 2\alpha }}{g}$
We get, by placing respective values,
${u^2} = \dfrac{{Rg}}{{\sin 2\alpha }} \\
\Rightarrow {u^2}= \dfrac{{16.7 \times 32.2}}{{\sin {{30}^ \circ }}}$
$ \therefore u \approx 32ft/\operatorname{s} $
As a result, the minimum speed at which the motorcycle can travel is $32ft/s$.
Note:The trajectory equation is the equation of projectile motion. So, if we know the x-component of an object's location, we can use the projectile motion equation to find the y-component of the position. The equation of the direction of projectile motion can be used to obtain all of the quantities involved in projectile motion, either directly or indirectly.
Complete step by step answer:
Given, Width of the ditch= $11.7ft$ and Length of the bike= $5ft$. The approach road makes an angle of ${15^ \circ }\left( \alpha \right)$ with the horizontal. The biker's total horizontal width that must be protected in order to safely cross the ditch,
$R = 11.7 + 5 = 16.7\,ft$
Acceleration due to gravity,
$a = g = 9.8\,m/s \\
= \,32.2\,ft/{s^2} \\ $
The horizontal range, as we know, is determined by
$R = \dfrac{{{u^2}\sin 2\alpha }}{g}$
We get, by placing respective values,
${u^2} = \dfrac{{Rg}}{{\sin 2\alpha }} \\
\Rightarrow {u^2}= \dfrac{{16.7 \times 32.2}}{{\sin {{30}^ \circ }}}$
$ \therefore u \approx 32ft/\operatorname{s} $
As a result, the minimum speed at which the motorcycle can travel is $32ft/s$.
Note:The trajectory equation is the equation of projectile motion. So, if we know the x-component of an object's location, we can use the projectile motion equation to find the y-component of the position. The equation of the direction of projectile motion can be used to obtain all of the quantities involved in projectile motion, either directly or indirectly.
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