Question

A skateboarder is moving 5.25 m/s when he starts to roll up a frictionless hill. How much higher is he when his velocity has reduced to 2.75 m/s. EXPLAIN HELP ME I DON'T GET THIS!!!!

Answer 1

Answer:

The height is 1.02 m.

Explanation:

Given that,

Speed of skateboarder = 5.25 m/s

Reduced velocity = 2.75 m/s

We need to calculate the height

Using mechanical energy,

Without frictional mechanical energy is same all points.

Using conservation of energy

$E_{i}=E_{f}$

$mgh+\dfrac{1}{2}mv_{i}^2=mgh+\dfrac{1}{2}mv_{f}^2$

When he starts, it means h = 0

So,

$\dfrac{1}{2}mv_{i}^2=mgh+\dfrac{1}{2}mv_{f}^2$

$\dfrac{1}{2}v_{i}^2=gh+\dfrac{1}{2}mv_{f}^2$

$h = \dfrac{v_{i}^2-v_{f}^2}{2g}$

$h=\dfrac{5.25^2-2.75^2}{2\times9.8}$

$h=1.02\ m$

Hence, The height is 1.02 m.

Answer 2

Mechanical energy E = mgh + 1/2mv²

When he starts, let h = 0 ⇒ E₁ = 1/2mv₁²
When he reaches height h ⇒ E₂ = mgh + 1/2mv₂²

Without friction, energy is conserved at all times.

E₁ = E₂
     ↓
1/2mv₁² = mgh + 1/2mv₂²
     ↓
1/2v₁² = gh + 1/2v₂²
     ↓
gh = 1/2(v₁² - v₂²)
    ↓
h = (v₁² - v₂²) / (2g)