Question

A skateboarder rolls horizontally off the top of a staircase and lands at the bottom of the stairs. The staircase has a horizontal length of 12.0 m, and the jump lasts 1.10 s. What is the skater's vertical velocity upon landing?

Answer 1

The skateboarder's vertical velocity upon landing is approximately 10.791 m/s downward, determined using the equation for uniformly accelerated motion under gravity.

Step-by-step explanation:

To determine the skateboarder's vertical velocity upon landing, we can use the physics principles of projectile motion. The motion of a projectile in the vertical direction is independent of its horizontal motion, and is influenced only by gravity (assuming no air resistance). Since the skateboarder rolls horizontally off the top, he starts with zero vertical velocity and the only acceleration acting on him is due to gravity (approximately 9.81 m/s2 downward). The vertical velocity upon landing can be calculated using the equation v = gt, where v is the final vertical velocity, g is the acceleration due to gravity, and t is the time spent in the air.

Upon substituting the values, we get v = (9.81 m/s2) × (1.10 s) = 10.791 m/s. Hence, the skateboarder's vertical velocity upon landing is approximately 10.791 m/s downward.