Question

A motorcycle daredevil is attempting to jump from one ramp onto another, as shown. The takeoff ramp makes an angle of 18.0o above the horizontal, and the landing ramp is identical. The cyclist leaves the ramp with a speed of 33.5 m/s. a) At the instant when the cyclist leaves the takeoff ramp, what are the horizontal AND vertical components of their velocity? B) What is the maximum distance that the landing ramp can be placed from the takeoff ramp so that the cyclist still lands on it?

Answer 1

The horizontal and vertical components of the daredevil's velocity are calculated using trigonometric functions based on the given launch speed and angle. The maximum distance for the ramp placement is found using the range equation of projectile motion.

Step-by-step explanation:

To solve for the horizontal and vertical components of the motorcycle daredevil's velocity, we use trigonometric functions. If the takeoff ramp angle is 18.0° and the speed is 33.5 m/s, the horizontal component (Vx) and the vertical component (Vy) can be found as follows:

Vx = V * cos(θ) = 33.5 m/s * cos(18.0°)

Vy = V * sin(θ) = 33.5 m/s * sin(18.0°)

To find the maximum distance the landing ramp can be placed for the daredevil to still land on it, we'll use the range equation for projectile motion, assuming that the landing ramp is at the same height as the takeoff ramp:

Range = (Vx * Vy) / (g/2), where g is the acceleration due to gravity (9.81 m/s²).

By calculating the components and substituting into the range equation, we can find the maximum distance.