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Setler [38]
3 years ago
7

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 red

uced to 2.75 m/s. EXPLAIN HELP ME I DON'T GET THIS!!!!
Physics
2 answers:
Lapatulllka [165]3 years ago
8 0

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.

Taya2010 [7]3 years ago
5 0
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)


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A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of
Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

h=ut-\dfrac{1}{2}gt^2

h=(u\sin60)\times t-\dfrac{1}{2}gt^2

Put the value into the formula

h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2

h=41.6\ m

(b). We need to calculate the maximum height of the ball

Using formula of height

h_{max}=\dfrac{(u\sin\theta)^2}{2g}

Put the value into the formula

h=\dfrac{(33\sin60)^2}{2\times 9.8}

h=41.67\ m

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

v=u-gt

v=u\sin\theta-gt

Put the value into the formula

v_{y}=33\times\sin 60-9.8\times3.0

v_{y}=-0.82\ m/s

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

v_{x}=u\cos\theta

Put the value into the formula

v_{x}=33\times\cos60

v_{x}=16.5\ m/s

We need to calculate the ball's impact speed

Using formula of velocity

v=\sqrt{v_{x}^2+v_{y}^2}

Put the value into the formula

v=\sqrt{(16.5)^2+(-0.82)^2}

v=16.52\ m/s

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

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Answer:

Option A

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Explanation:

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