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Katena32 [7]
2 years ago
14

a foul ball is hit into the stands at a baseball game. the ball rises to a height of 38 meters and is caught on its way down by

a fan standing in the bleaches 30 meters above the height at which it was hit
Physics
1 answer:
lisov135 [29]2 years ago
4 0

The velocity of the ball when it was caught is 12.52 m/s.

<em>"Your question is not complete it seems to be missing the following, information"</em>,

find the velocity of the ball when it was caught.

The given parameters;

maximum height above the ground reached by the ball, H = 38 m

height above the ground where the ball was caught, h = 30 m

The height traveled by the ball when it was caught is calculated as follows;

y = H - h

y = 38 - 30 = 8 m

The velocity of the ball when it was caught is calculated as;

v_f^2 = v_0 + 2gh\\\\v_f^2 = 0 + (2\times 9.8 \times 8)\\\\v_f^2 = 156.8\\\\v_f = \sqrt{156.8} \\\\v_f = 12.52 \ m/s

Thus, the velocity of the ball when it was caught is 12.52 m/s.

Learn more here: brainly.com/question/14582703

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Against the park ranger's advice, a visitor at a national park throws a stone horizontally off the edge of a 86 m high cliff and
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<h3>What is projectile?</h3>

When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion.

Time taken by the stone to reach the ground is

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The horizontal velocity is

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V(x) = 21.241 m/s

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A light flashes at position x=0m. One microsecond later, a light flashes at position x=1000m. In a second reference frame, movin
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Answer:

To the right relative to the original frame.

Explanation:

In first reference frame <em>S</em>,

Spatial interval of the event, \rm \Delta x=1000\ m-0\ m=1000\ m.

Temporal interval of the event, \rm \Delta t = 1\ \mu s=10^{-6}\ s.

In the second reference frame <em>S'</em>, the two flashes are simultaneous, which means that the temporal interval of the event in this frame is \rm \Delta t'=0\ s.

The speed of the frame <em>S' </em>with respect to frame <em>S</em> = v.

According to the Lorentz transformation,

\rm \Delta t'=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\left( \Delta t-\dfrac{v\Delta x}{c^2}\right ).\\\\Since,\ \Delta t'=0,\\\therefore \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\left( \Delta t-\dfrac{v\Delta x}{c^2}\right )=0\\\Rightarrow \Delta t-\dfrac{v\Delta x}{c^2}=0\\\dfrac{v\Delta x}{c^2}=\Delta t\\v=\dfrac{c^2\Delta t}{\Delta x }.\\\\Also, \ \Delta t,\ \Delta x>0\ \Rightarrow v>0.

And positive v means the velocity of the second frame<em> </em><em>S'</em> is along the positive x-axis direction, i.e., to the right direction relative to the original frame <em>S</em>.

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