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e-lub [12.9K]
3 years ago
5

A stone is thrown vertically upwards with a speed of 30.0 m/s.

Physics
1 answer:
matrenka [14]3 years ago
6 0

a)

Y₀ = initial position of the stone at the time of launch = 0 m

Y = final position of stone = 20.0 meters

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = ?

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

v² = 30² + 2 (- 9.8) (20 - 0)

v = 22.5 m/s


b)

Y₀ = initial position of the stone at the time of launch = 0 m

Y = maximum height gained

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = 0 m/s

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

0² = 30² + 2 (- 9.8) (Y - 0)

Y = 46 m



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Answer:

C. less than 950 N.

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C. less than 950 N

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Answer:

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Answer:

Vi = 8.28 m/s

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This problem is related to the projectile motion.

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Also we know that

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equate eq. 1 and eq. 2

5.3/t = Vi*0.5

5.3/0.5 = Vi*t

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The vertical distance is

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also we know that

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3 = 1.9 + Vi*0.866*t - 0.5gt²

3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²

3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²

3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²

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substitute Vi*t = 10.6 in above equation

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Vi*t = 10.6

Vi = 10.6/t

Vi = 10.6/1.28

Vi = 8.28 m/s

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