Question

Figure shows a 11.7 ft wide ditch with the approach roads at an angle of ${15}^0$ 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.

Answer 1

Given, Horizontal range,

x = 11.7 + 5

= 16.7 ft covered by the bike

g = 9.8 m/s = 32.2 ft/$s^2$

Now, R = $\frac{u^{2}sin2x}{g}$

$\Rightarrow u^2=\frac{Rg}{sin2x}=\frac{16.7\times 32.2}{1/2}$

$\Rightarrow$ u = 32 ft/s