Question

A copy-cat daredevil tries to reenact Evel Knievel's 1974 attempt to jump the Snake River Canyon in a rocket-powered motorcycle. The canyon is \(L=400 . \mathrm{m}\) wide, with the opposite rims at the same height. The height of the launch ramp at one rim of the canyon is \(h=8.00 \mathrm{~m}\) above the rim, and the angle of the end of the ramp is \(45.0^{\circ}\) with the horizontal. a) What is the minimum launch speed required for the daredevil to make it across the canyon? Neglect the air resistance and wind. b) Famous after his successful first jump, but still recovering from the injuries sustained in the crash caused by a strong bounce upon landing, the daredevil decides to jump again but to add a landing ramp with a slope that will match the angle of his velocity at landing. If the height of the landing ramp at the opposite rim is \(3.00 \mathrm{~m}\), what is the new required launch speed?

Short answer

Answer: Follow the step-by-step solution provided to calculate the minimum launch speed for part a and the new required launch speed for part b. The launch speed will vary based on the height of the landing ramp and the vertical displacement.

Step-by-step solution

(Step 1: Set up the known variables)

(In this case, we know the launch angle (\(45.0^{\circ}\)), the width of the canyon (400m), and the height of the launch ramp (8m). We also know the height of the landing ramp (3m) for part b of the problem.)

(Step 2: Determine the time of flight for part a)

(To solve this part, we need to determine the time it takes for the daredevil to reach the other side of the canyon. Calculate the time of flight using the horizontal component of the motion: $$ t = \frac{L}{v_x} $$ where \(L = 400\) m (width of the canyon), \(v_x\) is the horizontal component of the launch speed, and \(t\) is the time of flight. )

(Step 3: Determine the vertical component of velocity for part a)

(Now we will use the vertical component of the motion to determine the minimum required launch speed: $$ v_{yf}^2 = v_{yi}^2 + 2a(\Delta y) $$ where \(v_{yf}\) is the final vertical velocity (0 m/s because the opposite rims are at the same height), \(v_{yi}\) is the initial vertical velocity, \(a\) is the vertical acceleration (-9.81 m/s\(^2\)), and \(\Delta y\) is the vertical displacement. Since the launch ramp is 8m above the rim, the vertical displacement is \(-8\) m. )

(Step 4: Calculate the minimum launch speed for part a)

(Now we can use the relationship between the horizontal and vertical components of the velocity to calculate the minimum launch speed: $$ v = \sqrt{v_x^2 + v_{yi}^2} $$ Plug in the values obtained from Steps 2 and 3 to find the minimum launch speed for part a.)

(Step 5: Determine the time of flight for part b)

(For part b of the problem, we have a landing ramp with a height of 3m on the other side of the canyon. The time of flight will be the same as in part a because the width of the canyon remains the same. Therefore, we can use the same equation for time of flight that we derived in Step 2.)

(Step 6: Determine the vertical component of velocity for part b)

(For part b, the vertical displacement will be different since the landing ramp is now at a height of 3m. Therefore, \(\Delta y = -8 + 3 = -5\) m. Use the same equation for the vertical component of velocity as in Step 3 and substitute the new value of \(\Delta y\) to find the new final vertical velocity (\(v_{yf}\)).)

(Step 7: Calculate the new required launch speed for part b)

(Now we can use the same equation for the launch speed as in Step 4: $$ v = \sqrt{v_x^2 + v_{yi}^2} $$ Plug in the values obtained from Steps 5 and 6 to find the new required launch speed for part b.)

Key concepts

Kinematics

Kinematics is the branch of mechanics that describes the motion of objects without regard to the forces that cause the motion. It is foundational to understanding projectile motion, which is a form of two-dimensional motion where an object moves in a curved path under the influence of gravity.

In kinematics, key quantities such as displacement, velocity, acceleration, time, and projectile range are analyzed to predict the motion of an object. For example, initial velocity can be broken down into horizontal (\(v_x\)) and vertical (\(v_y\)) components when considering projectiles. The motion can be further described using equations that relate these quantities.

It's essential to understand the kinematic equations, which allow us to predict the object's position and velocity at any point in its trajectory. This understanding is applied to solve problems where we need to find launch speed or time of flight, as in our daredevil's jump across the canyon.

Trajectory

The trajectory of a projectile is the path it follows during its flight. In the case of our daredevil's rocket-powered motorcycle, the path would ideally be a parabola, which is typical for projectiles when air resistance is negligible.

The trajectory is influenced by the initial launch speed, launch angle (here, it's 45 degrees), and gravitational acceleration. The highest point in the trajectory is the peak, and the conditions at this point can be used to determine various aspects of the projectile's motion, such as the maximum height reached.

Ultimately, the trajectory is symmetrical for projectiles launched and landing at the same height, assuming no air resistance. This means the motorcycle must follow a trajectory that ensures it lands safely on the other side of the canyon, which is determined by both its launch speed and the angle of launch.

Launch Speed

Launch speed, or the initial velocity at which a projectile is propelled, is a crucial factor in determining the trajectory and ultimate success of a projectile's motion. For our daredevil, the minimum launch speed necessary to complete the canyon jump depends on both the distance to be covered and the difference in height between the launch and landing points.

Mathematically, the minimum speed can be found by breaking down the overall speed into its horizontal and vertical components. The horizontal component (\(v_x\)) determines how far the projectile will travel, while the vertical component (\(v_y\)) ensures it will reach a certain height.

For the jump with a symmetrical landing point at the same elevation as the takeoff point, these components will have equal magnitudes, since the launch angle is 45 degrees. However, when a landing ramp is added at a different height, the calculations adjust to ensure the trajectory meets this new point.

Time of Flight

Time of flight refers to the duration a projectile remains in the air from launch until it lands. For projectile motion where the launch and landing elevations are the same and air resistance is ignored, the time of flight can be determined exclusively by the horizontal motion. In our daredevil example, since gravity does not influence horizontal motion, the time of flight is based on the horizontal distance (\(L\text{, the width of the canyon}\)) and the horizontal component of the initial launch speed (\(v_x\text{, which remains constant during the flight}\)).

The time of flight is crucial in solving for the minimum launch speed, as it provides a link between the projectile's horizontal displacement and its velocity. Given that the horizontal distance the daredevil needs to cover remains constant, adjustments to the launch ramp height do not affect the time the motorcycle spends in the air, meaning that the time of flight for both cases in the problem would be the same.