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3 years ago

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a person dives off the edge of aa cliff 33m above the surface of the sea. assuming that air resistance is negligible, how long d

oes the dive last and with what speed does the person enter the water

1 answer:

USPshnik [31]3 years ago

3 0

Answer:

The time is $t = 2.595 \ s$

The speed is $v = 25.43 \ m/s$

Explanation:

From the question we are told that

The height of the cliff is $h = 33 \ m$

Generally from kinematic equation we have that

$h = ut + \frac{1}{2} gt^2$

before the jump the persons initial velocity is u = 0 m/s

So

$33 = 0 * t + \frac{1}{2} 9.8 * t^2$

=> $t = 2.595 \ s$

Generally from kinematic equation

$v= u + gt$

=> $v= 0 + 9.8 * 2.595$

=> $v = 25.43 \ m/s$

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