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: Identify values of v, i, and h

Given \( v = 32 \) (initial speed), \( i = 48 \) (initial height), and when the ball hits the ground, \( h = 0 \). Substitute into the height formula \( h=-16t^{2}-vt + i \).

$$ 0=-16t^{2}-32t + 48 $$

Step2: Simplify the equation

Divide the entire equation by -16 to simplify:

$$ 0 = t^{2}+2t - 3 $$

Step3: Factor the quadratic equation

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

$$ 0=(t + 3)(t - 1) $$

Step4: Solve for t

Set each factor equal to zero:

• \( t + 3 = 0 \) gives \( t=-3 \) (discard since time can't be negative)
• \( t - 1 = 0 \) gives \( t = 1 \)

Answer:

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