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Lostsunrise [7]
3 years ago
8

A 500-kg roller coaster car travels with some initial velocity along a track that is 5 m above the ground. The car goes down a s

mall hill and then coasts back up a higher hill. The second hill is 12 m above the ground. What must the initial velocity of the car be for the car to be going 3 m/s at the top of the second hill?
A.3.0 m/s
B.5.0 m/s
C.9.8 m/s
D.12 m/s
Physics
1 answer:
LUCKY_DIMON [66]3 years ago
4 0

Answer:

12m/s

Explanation:

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The horizontal distance covered by the ball before hitting the water is 70.4 m

Explanation:

The motion of the ball is the motion of a projectile, so it consists of two independent motions:

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We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:

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u = 23.5 m/s is the initial velocity

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Substituting,

u_y=(23.5)(sin 33.5^{\circ})=13.0 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity, downward

Substituting everything into the equation we get:

-16.5=13.0t-4.9t^2\\4.9t^2-13.0t-16.5=0

Solving the equation for t, we find the time of flight of the ball:

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t = 3.59 s

We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.

Now we analyze the horizontal motion of the ball. The horizontal velocity is constant and it is:

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Therefore, the horizontal distance covered in a time t is

d=v_x t

And substituting t = 3.59 s, we find

d=(19.6)(3.59)=70.4 m

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Learn more about projectile motion:

brainly.com/question/8751410

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<h3><u>Answer;</u></h3>

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<h3><u>Explanation</u>;</h3>
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