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olga55 [171]
3 years ago
8

A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin return

s to the juggler’s hand?
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
7 0

Answer:

1.68 s

Explanation:

From newton's equation of motion,

a = (v-u)/t.................................. Equation 1

Making t the subject of the equation

t =(v-u)g............................. Equation 2

Where t = time taken for the bowling pin to reach the maximum height, v = final velocity bowling pin, u = initial velocity of the bowling pin, g = acceleration due to gravity.

Note: Taking upward to be negative and down ward to be positive,

Given: v = 0 m/s ( at the maximum height), u = 8.20 m/s, g = -9.8 m/s²

t = (0-8.20)/-9.8

t = -8.20/-9.8

t = 0.84 s.

But,

T = 2t

Where T = time taken for the bowling pin to return to the juggler's hand.

T = 2(0.84)

T = 1.68 s.

T = 1.68 s

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The best I can do for you is something like this:

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The amount you'll change its momentum is called the <u><em>impulse</em></u> you give it.  The quantity of impulse is (force) x (length of time you push on it).

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Divide each side of that equation by (20,000 Newtons). Then it says:

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