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?

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?

Answer

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.