Final answer:
To calculate the height of the cliff, we analyze the horizontal motion to find the time in the air and then use this to find the vertical displacement. The cyclist is in the air for 0.922 seconds, which when plugged into the kinematic equation for vertical motion yields a cliff height of approximately 4.18 m.
Step-by-step explanation:
The student's question is asking to calculate the height of a cliff from which a cyclist has jumped with an initial velocity of 18 m/s, and lands at a distance of 16.6 m away. To determine the height of the cliff, we can use the kinematic equations for projectile motion. First, we need to separate the motion into horizontal and vertical components. Since the horizontal velocity (vx) remains constant, we can calculate the time (t) the cyclist is in the air using the horizontal distance (d) and horizontal component of the initial velocity (vx).
The horizontal component of the initial velocity is given by vx = vinitial × cos(θ) - where θ is the angle of the initial velocity vector above the horizontal. Since in our case the cyclist jumps off horizontally, θ equals to 0 degrees, so cos(θ) equals to 1. Hence, vx = vinitial. The time in the air is t = d / vx = 16.6 m / 18 m/s = 0.922 s.
Now we can calculate the vertical height (h) the cyclist falls using the vertical component of the initial velocity (vy), which is 0 m/s because the initial motion is purely horizontal, and the time (t) calculated before. We can use the kinematic equation h = vy × t + (1/2) × g × t², where g is the acceleration due to gravity (9.81 m/s²).
The vertical displacement is h = 0 × 0.922 s + (1/2) × (9.81 m/s²) × (0.922 s)² = 4.18 m. Therefore, the height of the cliff from which the cyclist jumped is approximately 4.18 m.