5- a ball is dropped from a very tall tower by a person, and the sound of it striking the earth is heard 3.2…

Explanation:

Step1: Let the height of the tower be $h$.

Let the time taken for the ball to fall be $t_1$ and the time taken for the sound to travel back be $t_2$. We know that $t_1 + t_2=3.2$ s. Also, for the free - fall of the ball, using the equation $h = \frac{1}{2}gt_1^{2}$ (where $g = 9.8$ m/s²), and for the sound travel, $h = v_s t_2$ (where $v_s=345$ m/s). So, $\frac{1}{2}gt_1^{2}=v_s(3.2 - t_1)$.

Step2: Substitute the values and solve the quadratic equation.

Substitute $g = 9.8$ m/s² and $v_s = 345$ m/s into $\frac{1}{2}\times9.8t_1^{2}=345\times(3.2 - t_1)$. $4.9t_1^{2}=1104-345t_1$. $4.9t_1^{2}+345t_1 - 1104 = 0$. Using the quadratic formula $t_1=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 4.9$, $b = 345$, $c=-1104$. $t_1=\frac{-345\pm\sqrt{345^{2}-4\times4.9\times(- 1104)}}{2\times4.9}=\frac{-345\pm\sqrt{119025 + 21638.4}}{9.8}=\frac{-345\pm\sqrt{140663.4}}{9.8}=\frac{-345\pm375}{9.8}$. We take the positive root $t_1=\frac{-345 + 375}{9.8}=\frac{30}{9.8}\approx3.06$ s.

Step3: Calculate the height of the tower.

Since $h=\frac{1}{2}gt_1^{2}$, substituting $g = 9.8$ m/s² and $t_1\approx3.06$ s. $h=\frac{1}{2}\times9.8\times(3.06)^{2}=4.9\times9.3636\approx46$ m.

Answer:

46 m