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3 years ago

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You throw a baseball directly upward at time ????=0 at an initial speed of 14.9 m/s. What is the maximum height the ball reaches

above where it leaves your hand? Ignore air resistance and take ????=9.80 m/s2.

1 answer:

xeze [42]3 years ago

8 0

Answer:

Maximum height, h = 11.32 meters

Explanation:

It is given that,

The baseball is thrown directly upward at time, t = 0

Initial speed of the baseball, u = 14.9 m/s

Ignoring the resistance in this case and using a = g = 9.8 m/s²

We have to find the maximum height the ball reaches above where it leaves your hand. Let the maximum height is h. Using third equation of motion as :

$v^2-u^2=2ah$

At maximum height, v = 0

and a = -g = -9.8 m/s²

$h=\dfrac{v^2-u^2}{2a}$

$h=\dfrac{0-(14.9\ m/s)^2}{2\times -9.8\ m/s^2}$

h = 11.32 meters

Hence, the maximum height of the baseball is 11.32 meters.

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Put the value into the formula

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$T_{2}=2\times300$

$T_{2}=600\ K$

Hence, The initial and final temperatures of the gas is 300 K and 600 K.

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3 years ago