Question

1. a ball is dropped from the top of a tower. it is seen to hit the ground 3.0 s after it was dropped. the height of the tower is (a) 44.1 m (b) 56.8 m (c) 45.7 (d) 9.8 m/s (e) none of the above

Explanation:

Step1: Identify the kinematic - equation

The ball is in free - fall, and the initial velocity $u = 0\ m/s$, the acceleration $a = g=9.8\ m/s^{2}$, and the time $t = 3.0\ s$. The kinematic equation for displacement $s$ is $s=ut+\frac{1}{2}at^{2}$.

Step2: Substitute the values

Since $u = 0\ m/s$, the equation simplifies to $s=\frac{1}{2}gt^{2}$. Substitute $g = 9.8\ m/s^{2}$ and $t = 3.0\ s$ into the equation: $s=\frac{1}{2}\times9.8\times(3.0)^{2}$.

Step3: Calculate the height

First, calculate $(3.0)^{2}=9.0$. Then, $\frac{1}{2}\times9.8\times9.0 = 4.9\times9.0=44.1\ m$.

Answer:

A. 44.1 m