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

To determine the minimum speed required for the motorbike to safely cross the ditch, we need to analyze the problem using the principles of projectile motion. You’ve already set up the problem well by identifying the key variables: the width of the ditch, the angle of projection, and the acceleration due to gravity. Let’s break this down step by step to find the solution.

Understanding the Problem

The motorbike leaves the road at the front when it starts its projectile motion. The width of the ditch is 11.7 ft, and the bike is projected at an angle of 15 degrees. The goal is to find the minimum speed (v) at which the bike must travel to ensure it crosses the ditch safely.

Key Variables

• Width of the ditch (R): 11.7 ft
• Angle of projection (θ): 15 degrees
• Acceleration due to gravity (g): 32 ft/s²
• Length of the bike: 5 ft (this will be considered when calculating the effective range)

Using the Range Formula

The formula for the horizontal range (R) of a projectile launched at an angle θ with an initial velocity v is given by:

R = (v² * sin(2θ)) / g

In this case, we can rearrange the formula to solve for v:

v² = (R * g) / sin(2θ)

Calculating the Required Velocity

Now, let’s plug in the values:

• R: 11.7 ft
• g: 32 ft/s²
• θ: 15 degrees

First, we need to calculate sin(2θ):

sin(2 * 15 degrees) = sin(30 degrees) = 0.5

Now substituting the values into the rearranged formula:

v² = (11.7 ft * 32 ft/s²) / 0.5

v² = (374.4 ft²/s²) / 0.5 = 748.8 ft²/s²

Taking the square root to find v:

v = √748.8 ft²/s² ≈ 27.4 ft/s

Considering the Length of the Bike

Since the bike is 5 ft long, we need to ensure that the back of the bike also clears the ditch. Therefore, the effective range we need to consider is:

Effective Range = Width of the ditch + Length of the bike = 11.7 ft + 5 ft = 16.7 ft

Now, we can recalculate the required velocity using the effective range:

v² = (16.7 ft * 32 ft/s²) / 0.5

v² = (534.4 ft²/s²) / 0.5 = 1068.8 ft²/s²

Taking the square root gives us:

v = √1068.8 ft²/s² ≈ 32.7 ft/s

Final Result

Thus, the minimum speed at which the motorbike should be moving on the road to safely cross the ditch is approximately 32.7 ft/s. This ensures that the entire length of the bike clears the ditch safely.