Question

Figure (3−E8) 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?
Figure
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.

Answer 1

At minimum speed of 32.7 ft/s, should a motorbike be moving on the road so, that it safely crosses the ditch.

Explanation:

Step 1:

Given data  :

                  Bike length = 5 ft

                  11.7 feet  wide ditch

In view of the horizontal range , x = 11.7 +5

                                                     x = 16.7 ft

                                                     θ = 15°

                                   acceleration = a = g = 9.8 m/s²

Conversion of m/s² to ft/s² :

                                                      a = g = 32.2 ft/s²

Step 2:

Now,

              $R=\frac{u^{2} \sin 2 \theta}{g}$

            $R g=u^{2} \sin 2 \theta$

          $\frac{R g}{\sin 2 \theta}=u^{2}$

             $u^{2}=\frac{R g}{\sin 2 \theta}$

             $u^{2}=\frac{R g}{\sin \left(2 \times 15^{\circ}\right)}$

             $u^{2}=\frac{R g}{\sin \left(30^{\circ}\right)}$

             $u^{2}=\frac{16.7 \times 32.2}{\sin \left(30^{\circ}\right)}$

             $u^{2}=\frac{537.74}{\frac{1}{2}}$

             $u^{2}=537.74 \times 2$

             $u^{2}=1075.48$  

Step 3:

Applying squaring on both sides we get the following,

               $u=\sqrt{1075.48}$

               $u=32.79 \mathrm{ft} / \mathrm{s}$

Thus, the motorcycle should be traveling on the road at 32.7 ft/s speed so, that it can safely cross the ditch .

Answer 2

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32.7 ft/s

hope it help