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

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
$R=\frac{u^2\sin 2\alpha }{g}$
By putting respective values, we get:
$$\begin{gathered} u^2=\frac{Rg}{\sin 2\alpha }=\frac{16.7\times 32.2}{\sin 30°} \\ \Rightarrow u\approx 32\mathrm{ft}/\mathrm{s} \end{gathered}$$

Therefore, the minimum speed with which the motorbike should be moving is 32 ft/s.