Question

A pebble is thrown upward from the edge of a building 132 feet above the ground with an initial upward velocity of 4 feet per second. How long does it take to reach the ground? (\(\begin{array}{llll}\text { (A) } 2 \frac{1}{2} \text { seconds } & \text { (B) } 2 \frac{3}{4} \text { seconds } & \text { (C) } 3 \text { seconds } & \text { (D } 6 \text { seconds }\end{array}\)

Short answer

The pebble takes approximately 3 seconds to reach the ground. So, the correct answer option is (C) 3 seconds.

Step-by-step solution

Understand & Define Variables

The pebble has been thrown up, with an initial velocity \(u = 4 ft/s\). It is thrown from a height of \(S = 132ft\). Since the pebble is moving under gravity, acceleration \(a\) will be equal to -9.8 m/s\(^2\), or approximately \(a = -32 ft/s^2\) (converted from meters to feet). We're trying to find the time \(t\) it takes for the pebble to reach the ground.

Set Up the Equation

Now, we'll use the motion under gravity equation: \(S = ut + \frac{1}{2}at^2\). We can fill in the variables: \(132 = 4t + \frac{1}{2}(-32)t^2\). This simplifies to: \(0 = -16t^2 + 4t + 132\).

Solve for the time \(t\)

We can solve the quadratic equation for \(t\) (Remember to only consider the positive value for time). This can be done using the quadratic formula: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=-16\), \(b=4\), and \(c=132\). Substituting these values, we find \(t \approx 3 seconds\).