Question

A skateboarder rolls off a 2.5 m high bridge into the river. If the skateboarder was originally moving at 7.0 m/s, how much time did the skateboarder spend in the air and how far will be his final position from the bridge?

Answer 1

The skateboarder spends approximately 0.716 seconds in the air and will land about 5.01 meters away from the bridge.

Step-by-step explanation:

To determine how long the skateboarder spends in the air and how far away from the bridge he ends up, we have to split the problem into vertical and horizontal components.

Vertical Motion

Using the kinematic equation for vertical motion y = v_{0y}t + 0.5gt^2, where g is the acceleration due to gravity (9.8 m/s2), y is the vertical displacement (2.5 m), and v_{0y} is the initial vertical velocity (0 m/s since he rolls off horizontally).

Solving for time t this becomes 2.5 m = 0.5 * 9.8 m/s2 * t^2. Simplifying this we find t to be approximately 0.716 seconds, which is the time the skateboarder spends in the air.

Horizontal Motion

The horizontal distance the skateboarder travels is simply the horizontal velocity (7.0 m/s) multiplied by the time in the air. So, the horizontal displacement x is x = v_{0x} * t, which gives us x = 7.0 m/s * 0.716 s which is approximately 5.01 meters.

Therefore, the skateboarder ends up approximately 5.01 meters from the bridge after falling in the air for about 0.716 seconds.

Answer 2

t = 0.714 s and x = 5.0 m

Step-by-step explanation:

This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed

vₓ = 7.0 m / s

Let's find the time it takes to get to the river

y = y₀ + v_{oy} t - ½ g t²

the initial vertical speed is zero and when it reaches the river its height is zero

0 = y₀ + 0 - ½ g t²

t =
`\sqrt{(2y_o)/(g) }`

t = ra 2 2.5 / 9.8

t = 0.714 s

the distance traveled is

x = vₓ t

x = 7.0 0.714

x = 5.0 m