Question

Q4 You throw a ball upward. Its height h, in feet, after t seconds can be modeled by the function h= -16t2+30t+6 After how many seconds will it hit the ground? (Use the Quadratic Formula)

Answer 1

To find the time at which the ball hits the ground, we need to solve the equation h = 0. In this case, the equation becomes:

0 = -16t^2 + 30t + 6

To solve this quadratic equation, we can use the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:

x = (-b ± √(b^2 - 4ac))/(2a)

In our case, a = -16, b = 30, and c = 6. Substituting these values into the quadratic formula, we have:

t = (-30 ± √(30^2 - 4(-16)(6)))/(2*(-16))

Simplifying further:

t = (-30 ± √(900 + 384))/(2*(-16))
t = (-30 ± √1284)/(2*(-16))

Taking the square root of 1284:

√1284 ≈ 35.83

t = (-30 ± 35.83)/(2*(-16))

Now we can solve for the two possible values of t:

t₁ = (-30 + 35.83)/(2*(-16)) ≈ 0.42
t₂ = (-30 - 35.83)/(2*(-16)) ≈ 3.79

Since time cannot be negative, we discard the negative solution t₂. Therefore, the ball will hit the ground after approximately 0.42 seconds.