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/s^{2}

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.