A BMX bike jumps off a 10 meter tall ramp with an angle of inclination of 35 degrees, and a speed of 12 m/s.

Question

A BMX bike jumps off a 10 meter tall ramp with an angle of inclination of 35 degrees, as shown below, and a speed of 12 m/s.

The landing ramp is 6 m high, and 18.2 m away from the base of the launching ramp. find the angle (to one decimal) of inclination of the landing ramp, so that the bike hits the ramp, traveling in the same direction as the ramp slope.

Answer 1

Hello Quinn:

To determine the required angle of inclination of the landing ramp, we need to determine the direction of the velocity vector when the bike reaches the landing ramp. We can analyze the situation as projectile motion, and use basic kinematic equations in both the x- and y- directions. The given initial conditions are:

Initial Position:

x_o = 0

y_o = 10 m

Initial Speed:

v_o = 12m/s

Angle of inclination of the launch ramp:

θ = 35°

The x- and y-components of the initial velocity are

V_o,x = V_ocos(θ) = 12cos(35) = 9.83m/s

V_o,y = V_osin(θ) = 12sin(35) = 6.88m/s

Use x = x_o + V_o,x t to find the time it takes the bike to reach the landing ramp. (Set x = 18.2 m).

18.2m = (9.83m/s) t,

so t = 1.85s

The x-component of the velocity when the bike lands is V_x = 9.83m/s. (We assume there is no acceleration in the x-direction, so V_x is constant.)

The y component of the velocity when the bike lands is given by V_y = V_o,y - gt

That is,

V_y = 6.88m/s - (9/8m/s²)1.85s = -11.26m/s

Let β be the angle for the velocity when the bike lands. Then

tan(β) = V_y / V_x = (-11.26m/s)/(9.83m/s). This gives

β= -48.9° (The negative indicates the angle is clockwise relative to the positive x-axis.)

We give the angle of inclination for the landing ramp as a positive angle, 48.9°.

Let me know if you need any more help.

William