Question

1. The drawing shows a skateboarder moving at 5.4 m/s along a horizontal section of a track that is
slanted upward by 48° above the horizontal at its end, which is 0.40 m above the ground. When she
leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air
resistance, find the maximum height H to which she rises above the end of the track.

Answer 1

Projectile motion is the curved path motion of a body launched into the air near the Earth's surface and having a horizontal velocity.

• The maximum height to which the skateboarder rises, H, is approximately  0.6 meters above the end of the track.

Reason:

Given parameter are;

Initial velocity of the skateboarder, v₁ = 5.4 m/s

Inclination of the track, above the horizontal, θ = 48°

Height of the end of the elevated track, h ≈ 0.40 m

Path of the skateboarder when she leaves the track = Path of a projectile

Required:

Maximum height H to which she rises above the end of the track.

Solution;

From v² = u² - 2·g·h, at the end of the track where;

h = 0.40 m

u = Initial velocity = 5.4

g = 9.81 m/s²

We have;

v₂² ≈ 5.4² - 2 × 9.81 × 0.40 = 21.312

• The velocity at which she leaves the track, v₂ ≈ √(21.312 m²/s²).

At the maximum height, H, we have;

$\displaystyle v__y$ = 0

Therefore, from $\displaystyle v__y$² = $\displaystyle u__y$² - 2·g·H, where;

$\displaystyle u__y$² = 2·g·H

• $H = \dfrac{u_y^2}{2 \cdot g}$

Which gives;

• $H = \dfrac{21.312}{2 \times 9.81} \times sin^2(48^{\circ}) \approx 0.6$

Therefore;

• The maximum height to which she rises above the end of the track, H ≈ 0.6 m.

Learn more about projectile motion here:

https://brainly.com/question/12125940