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3 years ago

9

In a bizarre but harmless Superman fell from the top of the Eiffel How fast was Superman traveling when he hit the ground 7.80 s

econds after falling ?

1 answer:

brilliants [131]3 years ago

5 0

Vf = final velocity
Vi = initial velocity
g = acceleration due to gravity
t = time

Vf = Vi + gt

Vf = 0 + (9.8)(7.80)
Vf = 76.44 m/s

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The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

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The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

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$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

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Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

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