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3 years ago

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A ball is thrown upward with an initial velocity of 48 ft/s from a height 864 ft. h = -16t +48t +864 After how many seconds will

the ball reach the ground? (i.e. For what positive value of t is h=0?) ...?

1 answer:

neonofarm [45]3 years ago

6 0

0 = -16 t^2 + 48t + 864 (Divide all by 16)

-t^2 + 3t + 54

(t-9) (t+6) = 0

t = 9s
t = -6s

so the positive value would be : 9s

hope this helps

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Answer:

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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:

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

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

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

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

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

$\mu = 0.859$

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

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