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2 years ago

11

A lacrosse ball that is thrown straight upwards reaches a maximum height of 4.5 m. At what initial velocity was it thrown? (note

: final velocity, v = 0, at the top).

1 answer:

shtirl [24]2 years ago

5 0

Answer:

The initial velocity was 9.39 m/s

Explanation:

<em>Lets explain how to solve the problem</em>

The ball is thrown straight upward with initial velocity u

The ball reaches a maximum height of 4.5 m

At the maximum height velocity v = 0

The acceleration of gravity is -9.8 m/s²

We need to find the initial velocity

The best rule to find the initial velocity is <em>v² = u² + 2ah</em>, where v is

the final velocity, u is the initial velocity, a is the acceleration of

gravity and h is the height

⇒ v = 0 , h = 4.5 m , a = -9.8 m/s²

⇒ 0 = u² + 2(-9.8)(4.5)

⇒ 0 = u² - 88.2

Add 88.2 to both sides

⇒ 88.2 = u²

Take square root for both sides

⇒ u = 9.39 m/s

<em>The initial velocity was 9.39 m/s</em>

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Answer:

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Explanation:

The two cars collide each other inelastically, then we can determine the resulting velocity by the Principle of Momentum Conservation:

$m_{A}\cdot v_{A} + m_{B}\cdot v_{B} = (m_{A} + m_{B})\cdot v$ (1)

Where:

$m_{A}$ , $m_{B}$ - Masses of the cars, in kilograms.

$v_{A}$ , $v_{B}$ - Initial velocities of the cars, in meters per second.

$v$ - Velocity of the resulting system, in meters per second.

If we know that $m_{A} = 2120\,kg$ , $v_{A} = 13.4\,\frac{m}{s }$ , $m_{B} = 2810\,kg$ and $v_{B} = 0\,\frac{m}{s}$ , then the velocity of the resulting system:

$v = \frac{m_{A}\cdot v_{A}+m_{B}\cdot v_{B}}{m_{A}+m_{B}}$

$v = \frac{(2120\,kg)\cdot \left(13.4\,\frac{m}{s} \right)+(2810\,kg)\cdot \left(0\,\frac{m}{s} \right)}{2120\,kg + 2810\,kg}$

$v = 5.762\,\frac{m}{s}$

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$K = W_{f}$ (2)

$\frac{1}{2}\cdot (m_{A}+m_{B})\cdot v^{2} = \mu\cdot (m_{A}+m_{B})\cdot g \cdot s$

$\frac{1}{2}\cdot v^{2} = \mu \cdot g\cdot s$ (2b)

Where:

$\mu$ - Coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$s$ - Travelled distance, in meters.

If we know that $v = 5.762\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $s = 1.97\,m$ , then the coefficient of friction is:

$\mu = \frac{v^{2}}{2\cdot g\cdot s}$

$\mu = \frac{\left(5.762\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.97\,m)}$

$\mu = 0.859$

The coefficient of friction between the cars and the road is 0.859.

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