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blsea [12.9K]
3 years ago
5

A roller coaster is released from the top of a track that is 125 m high. Find the rollar coaster speed when it reaches ground le

vel.​
Physics
1 answer:
astra-53 [7]3 years ago
6 0

Explanation:

A roller coaster is released from the top of a track that is 125 m high. Find the roller coaster speed when it reaches ground level. 75. A 1500 kg car, moving at a speed of 20 m/s comes to a halt. How much work was done .

<h3><u>HOPE </u><u>IT </u><u>HELPS</u><u /><u /></h3>

<h2><em>mark </em><em>me </em><em>in </em><em>brainliest </em><em>answers </em><em>please </em><em>please </em><em>please </em><em /><em /><em /><em /><em /><em /><em /><em /><em /><em /></h2>
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You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration
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Answer: 3.41 s

Explanation:

Assuming the question is to find the time t the ball is in air, we can use the following equation:

y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}

Where:

y=0m is the final height of the ball

y_{o}=40 m is the initial height of the ball

V_{o}=10 m/s is the initial velocity of the ball

t is the time the ball is in air

g=9.8 m/s^{2} is the acceleration due to gravity  

\theta=30\°

Then:

0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}

0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}

Multiplying both sides of the equation by -1 and rearranging:

4.9 m/s^{2}t^{2}-5m/s t-40 m=0

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t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

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a=4.9

b=-5

c=-40

Substituting the known values:

t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}

Solving the equation and choosing the positive result we have:

t=3.41 s  This is the time the ball is in air

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