Question

1. an object is thrown from a height of 120 meters with an upward velocity of 24 meters per second.

a) write an equation to model the height of the object at any time.
a)________________________
b) how long does it take the object to reach the ground? show all work. pythagorean theorem? 2 points ec

Explanation:

Step1: Recall the height - time formula

The general formula for the height \(h(t)\) of an object in vertical - motion under the influence of gravity is \(h(t)=- 4.9t^{2}+v_{0}t + h_{0}\), where \(v_{0}\) is the initial velocity and \(h_{0}\) is the initial height. Given \(v_{0} = 24\) m/s and \(h_{0}=120\) m.
So the equation is \(h(t)=-4.9t^{2}+24t + 120\)

Step2: Find the time when the object reaches the ground

When the object reaches the ground, \(h(t)=0\). So we need to solve the quadratic equation \(-4.9t^{2}+24t + 120 = 0\).
The quadratic formula for \(ax^{2}+bx + c = 0\) is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a=-4.9\), \(b = 24\), and \(c = 120\)

First, calculate the discriminant \(\Delta=b^{2}-4ac=(24)^{2}-4\times(-4.9)\times120\)
\(=576+2352\)
\(=2928\)

Then \(t=\frac{-24\pm\sqrt{2928}}{2\times(-4.9)}=\frac{-24\pm54.11}{-9.8}\)

We have two solutions for \(t\):
\(t_{1}=\frac{-24 + 54.11}{-9.8}=\frac{30.11}{-9.8}\approx - 3.07\) (rejected since time cannot be negative)
\(t_{2}=\frac{-24-54.11}{-9.8}=\frac{-78.11}{-9.8}\approx7.97\)

Answer:

a) \(h(t)=-4.9t^{2}+24t + 120\)
b) It takes approximately \(t = 8.0\) seconds for the object to reach the ground.