a ball is dropped from a height of 48 feet with an initial speed of 32 feet per second. the height of the…

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:?

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:?

Answer

Explanation:

Step1: Simplify the quadratic 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) gives (t=-3)
  • (t - 1 = 0) gives (t = 1)

Since time cannot be negative, we discard (t=-3).

Answer:

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