Question

plss solve this ques


hurry up

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Answer 1

Correct Question :

A 11.7 ft wide ditch with the approach roads at and angle of 15°with the horizontal. With what minimum speed should a mororbike be moving on the road so that it safely croses the ditch?

Assume that the length of thebike is 5 ft, and it leaves the road when the front part runs out of the approch road.

Theory :

Equation of trajectory:

$y =x \tan( \theta) - \frac{gx {}^{2} }{2u {}^{2} \cos {}^{2} ( \theta) }$

•Given :

width of ditch =11.7 ft

length of bike =5ft

θ=15°

g=32.2 ft/s²

Solution:

Total distance covered (x )

= ditch + length of bike

= (11.7+5)ft = 16.7ft

Now use equation of trajectory:

$y =x \tan( \theta) - \frac{gx {}^{2} }{2u {}^{2} \cos {}^{2} ( \theta) }$

since distance travel in y =0

$\implies 0 =x \tan( \theta) - \frac{gx {}^{2} }{2u {}^{2} \cos( \theta) }$

$\implies \: x \tan( \theta) = \frac{gx {}^{2} }{2u {}^{2} \cos( \theta) }$

$\implies x \frac{ \sin( \theta) }{ \ \cos( \theta) } = \frac{gx {}^{2} }{2u {}^{2} \cos {}^{2} ( \theta) }$

$\implies \: x \sin( \theta) = \frac{gx {}^{2} }{2u {}^{2} \cos( \theta) }$

$\implies \: u = \sqrt{ \frac{gx }{ \sin(2 \theta) } }$

put the values

$\implies u = \sqrt{ \frac{32.2 \times (16.7) }{ \frac{1}{2} } }$

$\implies \: u = \sqrt{1075.48}$

$\implies \: u = 32 \: ft s {}^{ - 1} \: (approx)$

Therefore,

The minimum speed should a mororbike be moving on the road so that it safely croses the ditch is 32 ft/s.