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zloy xaker [14]

1 year ago

5

A ball is thrown upwards. Neglecting air resistance, what initial upward speed does the ball need to remain in the air for a tot

al time of 10 seconds

1 answer:

melamori03 [73]1 year ago

5 0

Answer:

H = 1/2 g t^2 time to reach top of trajectory

v = g t time to reach top of trajectory when v is initial speed upwards

v = 5 g = 49 m/s 5 sec upwards and 5 sec downwards

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Explanation:

Given that,

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$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s$

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Radial component of acceleration,

$a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2$

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