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?

Explanation:

Step1: Substitute values into the height formula

We know that \( v = 32 \) (initial speed) and \( i = 48 \) (initial height), and when the ball hits the ground, \( h = 0 \). So substitute these values into the formula \( h=- 16t^{2}-vt + i\), we get:
\(0=-16t^{2}-32t + 48\)

Step2: Simplify the quadratic equation

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

Step3: 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:
\(t^{2}+2t - 3=(t + 3)(t - 1)=0\)

Step4: Solve for t

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

Answer:

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