Question

a ball is dropped from a height of 48 feet with an initial speed of 32 feet per second. the height of the ball is $h = -16t^2 - vt + i$ in feet, where $v$ is the initial speed and $i$ is the initial height. when will the ball hit the ground? equation: $-16t^2 - 32t + 48 = 0$ solution:?

Explanation:

Step1: Simplify the equation

Divide the entire equation \(-16t^{2}-32t + 48 = 0\) by \(-16\) to simplify.
\(\frac{-16t^{2}}{-16}-\frac{32t}{-16}+\frac{48}{-16}=\frac{0}{-16}\)
Which gives \(t^{2}+ 2t - 3=0\)

Step2: Factor the quadratic equation

Factor \(t^{2}+2t - 3\). We need two numbers that multiply to \(-3\) and add to \(2\). The numbers are \(3\) and \(-1\).
So, \(t^{2}+2t - 3=(t + 3)(t - 1)=0\)

Step3: Solve for t

Set each factor equal to zero:
\(t + 3 = 0\) or \(t - 1 = 0\)
For \(t + 3 = 0\), we get \(t=-3\). For \(t - 1 = 0\), we get \(t = 1\). Since time cannot be negative in this context, we discard \(t=-3\).

Answer:

The ball hits the ground at \(t = 1\) second.