As shown in the figure, a ditch is 11.7 ft wide and has approach roads inclined at an angle of 15^ with the horizontal. Determine the minimum speed in ft/s at which a motorbike must travel on the road to safely cross the ditch. Assume the bike has a length of 5 ft and that it leaves the road exactly when its front part runs out of the approach road.

Step-by-step solution

The bike acts as a projectile starting the moment its front wheel reaches the edge. At this instant, the bike is on the incline, so the horizontal distance of its back wheel from the edge is 5 15^ . To safely cross, the back of the bike must clear the ditch. Thus, the total required horizontal range is: R = 11.7 + 5 15^ R 11.7 + 5(0.966) = 11.7 + 4.83 = 16.53 ft Using the horizontal range formula R = u^2 2 g with g = 32 ft/s ^2 and = 15^ : 16.53 = u^2 30^ 32 16.53 = u^2 (0.5) 32 u^2 = 16.53 32 0.5 = 1057.92 u = 105