Figure shows a 11.7 ft wide ditch with the approach roads at an angle of 15 degrees 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 approach road. For figure see H.C.V. Concepts of physics part 1 page 53. Here’s my solution Let the velocity with which the projectile motion of bike starts for crossing the ditch be v ft/s. Horizontal range(R)=11.7ft.Angle of projection(a)=15 degrees(corresponding angles)Acceleration due to gravity(g)=32ft/s^2So, by the formula of horizontal range(v^2 x sin(2 x 15)degrees )/32=11.7v^2=748.8 I am not able to approach further what should I do
Figure shows a 11.7 ft wide ditch with the approach roads at an angle of 15 degrees 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 approach road. For figure see H.C.V. Concepts of physics part 1 page 53.
Here’s my solution
Let the velocity with which the projectile motion of bike starts for crossing the ditch be v ft/s.
Horizontal range(R)=11.7ft.
Angle of projection(a)=15 degrees(corresponding angles)
Acceleration due to gravity(g)=32ft/s^2
So, by the formula of horizontal range
(v^2 x sin(2 x 15)degrees )/32=11.7
v^2=748.8 I am not able to approach further what should I do
Pritesh Ramya , 9 Years ago
Grade 11
