# Answer in Physics for Fel #163443

A book is thrown straight upward from the top of the MASH building with an initial speed of 14m/s. If the building is 92m tall, A).how long is the book in the air?B). What velocity will the book hit the ground with? Ignore air resistance.

A) The speed of the book until it reaches its highest point is given as:

“v = v_0-gt”

where “v_0 = 14m/s” is the initial speed, and “g = 9.81 m/s^2” is the gravitational acceleration. At the highest point “v = 0”, thus, the time required to reach it is:

“0 = v_0-gt_h\nt_h = dfrac{v_0}{g}”

The time to return to the initial level of launching (top of the building) is:

“t_b = 2t_h = dfrac{2v_0}{g}\nt_b = dfrac{2cdot 14}{9.81} approx 2.85 s”

At this time the book will have the same speed “v_0”, but now directed downward (this is so, according to the energy conservation law). After it passes the top of the building, its traveled distance is given as:

“h = v_0t + dfrac{gt^2}{2}”

Book lands when “h = 92m” (the height of the boulding). Substituting this value into the eqation above and solving for “t”, obtain the time that was took to fall from the top of the building to ground:

“4.905t_g^2 + 14t_g-92 = 0\nt_g approx 3.13s”

Thus, the total time in the air is:

“t = t_b + t_g = 2.85s + 3.13s = 5.98s”

B) The velocity of the book after it passes the top of the building is given as follows:

“v = v_0 + gt”

Substituting “t = t_g”, obtain:

“v = 14m/s + 9.81m/s^2cdot 3.13s approx 44.7 m/s”

Answer. A) 5.98 s, B) 44.7 m/s.

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