Question

question 30
a ball is dropped from rest from a tower and strikes the ground 110 m below. approximately how many seconds does it take the ball to strike the ground after being dropped?
3.50 s
2.50 s
16.0 s
4.74 s
12.5 s

Explanation:

Step1: Identify the kinematic equation

The kinematic equation for vertical motion under gravity when the initial velocity \(u = 0\) is \(h=ut+\frac{1}{2}gt^{2}\). Since \(u = 0\), the equation simplifies to \(h=\frac{1}{2}gt^{2}\), where \(h\) is the height, \(g = 9.8\ m/s^{2}\) (acceleration due to gravity), and \(t\) is the time.

Step2: Solve for \(t\)

From \(h=\frac{1}{2}gt^{2}\), we can express \(t\) as \(t=\sqrt{\frac{2h}{g}}\).
Substitute \(h = 110\ m\) and \(g=9.8\ m/s^{2}\) into the formula:

$$t=\sqrt{\frac{2\times110}{9.8}}=\sqrt{\frac{220}{9.8}}\approx\sqrt{22.45}\approx4.74\ s$$

Answer:

\(4.74\ s\) (the fourth option)