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2 years ago

50 points!!! Kinetics

Physics

2 answers:

enyata [817]2 years ago

7 0

Answer:

h = 98m

Explanation:

The ball follows a projectile motion.

It it's at the same height at t = 2 and t =10,

It was at a max height at:

t = (2+10)/2

= 12/2

= 6s

Conventional: upwards positive

s = ut + ½at²

s = ut + ½(-9.8)t²

s = ut - 4.9t²

s is equal at t = 2 and t = 10

2u - 4.9(2)² = 10u - 4.9(10)²

2u - 19.6 = 10u - 490

8u = 470.4

u = 58.8

at t = 2

s = 58.8(2) - 4.9(2)²

s = 98 m

iris [78.8K]2 years ago

3 0

Answer:

98 m √

Explanation:

How about s = Vo * t + ½at² ?

s = h = Vo * 2s - 4.9m/s² * (2s)² = 2Vo - 19.6

and

h = Vo * 10s - 4.9m/s² * (10s)² = 10Vo - 490

Subtract 2nd from first:

0 = -8Vo + 470.4

Vo = 58.8 m/s

h = 58.8m/s * 2s - 4.9m/s² * (2s)² = 98 m

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3 years ago

The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total

pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass $m$ travelling at a velocity $\vec{v}$ . The momentum $\vec{p}$ of this object would be:

$\vec{p} = m \cdot \vec{v}$ .

For the law of conservation of momentum, consider two objects: object $\rm a$ and object $\rm b$ . Assume that these two objects collided with each other.

Let $m_{\rm a}$ and $m_{\rm b}$ denote the mass of the two objects.
Let $\vec{v}_{\rm a}(\text{initial})$ and $\vec{v}_{\rm b}(\text{initial})$ denote the velocity of the two object right before the interaction.
Let $\vec{v}_{\rm a}(\text{final})$ and $\vec{v}_{\rm b}(\text{final})$ denote the velocity of the two objects right after the interaction.

The momentum of the two objects right before the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ , respectively.
The momentum of the two objects right after the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ , respectively.

The sum of the momentum of the two objects would be:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ right before the collision, and
$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ .