Question

an object is moving straight up only under the influence of gravity, and experiencing zero air drag. when it is at a height of 185 meters above the ground, it is moving up at a speed of 5 meters per second. find the time in seconds that the object has left before it hits the ground. provide at least one decimal place

Explanation:

Step1: Identify the kinematic equation

Use the equation \(y = y_0 + v_0t-\frac{1}{2}gt^2\). Here, \(y = 0\) (ground level), \(y_0=185\) m, \(v_0 = 5\) m/s, and \(g = 9.8\) m/s².

Step2: Substitute values into the equation

\(0=185 + 5t-\frac{1}{2}\times9.8t^2\). Simplify to \(4.9t^2-5t - 185=0\).

Step3: Use the quadratic formula

For \(ax^2+bx + c = 0\), \(t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 4.9\), \(b=- 5\), \(c=-185\).
\(t=\frac{5\pm\sqrt{(-5)^2-4\times4.9\times(-185)}}{2\times4.9}=\frac{5\pm\sqrt{25 + 3626}}{9.8}=\frac{5\pm\sqrt{3651}}{9.8}\).
\(\sqrt{3651}\approx60.42\). So \(t=\frac{5\pm60.42}{9.8}\).
We take the positive root \(t=\frac{5 + 60.42}{9.8}\) (since time can't be negative).

Step4: Calculate the value

\(t=\frac{65.42}{9.8}\approx6.7\) s.

Answer:

\(6.7\)