Answer to Question #163443 in Physics for Fel

Question

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.

Answer 1

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:

$$0 = v_0-gt_h\\ t_h = \dfrac{v_0}{g}$$

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

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

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:

$$4.905t_g^2 + 14t_g-92 = 0\\ t_g \approx 3.13s$$

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:

$$v = 14m/s + 9.81m/s^2\cdot 3.13s \approx 44.7 m/s$$

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