Question

A bike rider going over a ramp. The rider’s speed at the top of the ramp is 10 m/s. The angle between the ramp and the groundis30°.The top of the ramp is1.1m above the ground. a)The vertical velocity of the rider just as they leave the top of the ramp is 5 m s–1.Calculate the maximum height that the rider will reach above the ground.

Answer 1

Find the vertical velocity of the rider just as they leave the top of the ramp:

Vertical component of velocity = 5 m/s

Horizontal component of velocity = 10 m/s

Total velocity = √(5² + 10²) = √125 ≈ 11.2 m/s

Calculate the time it takes for the rider to reach the maximum height:

Initial vertical velocity = 5 m/s

Final vertical velocity = 0 m/s

Acceleration due to gravity = -9.81 m/s²

Using the kinematic equation vf = vo + at, where vf = 0, vo = 5 m/s, and a = -9.81 m/s²:

t = (vf - vo) / a = (0 - 5) / (-9.81) ≈ 0.51 s

Calculate the maximum height that the rider will reach above the ground:

Using the kinematic equation d = vot + 1/2at², where d is the maximum height above the ground, vo = 5 m/s, a = -9.81 m/s², and t ≈ 0.51 s:

d = 5(0.51) + 1/2(-9.81)(0.51)² ≈ 1.52 m

Therefore, the maximum height that the rider will reach above the ground is approximately 1.52 meters.