Question

Need help understanding the formula

Answer 1

For this case we have the following equation:

$H (t) = -16t ^ 2 + vt + s$

Where,

t: time

v: initial speed

s: initial height

Substituting values we have:

$H (t) = -16t ^ 2 + 15t + 4$

By the time the ball hits the ground we have:

$-16t ^ 2 + 15t + 4 = 0$

From here, we clear the value of time:

Using the quadratic equation we have:

$t=\frac{-15+/-\sqrt{15^2-4(-16)(4)}}{2(-16)}$

Doing the calculations we have:

$t1=-0.217\\t2=1.154$

Discarding the negative root, we have that the time is:

$t=1.154$

Answer:

it takes the ball to hit the floor about:

1.15 seconds

Answer 2

v is the initial speed and s is the initial height so the equation is

$h(t) = -16t^2+15t+4$

Set h(t) equal to 0 and solve for t

$-16t^2+15t+4=0 \\a=-16,b=15,c=4 \\x_{1,2}= \frac{-b \pm \sqrt{b^2-4ac}}{2a}= \frac{-15 \pm \sqrt{15^2-4(-16)4}}{2(-16)}=\frac{-15 \pm \sqrt{225+256}}{-32}=\frac{-15 \pm \sqrt{481}}{-32} \\x_1=\frac{-15 + \sqrt{481}}{-32} \approx -0.22 \\x_1=\frac{-15 - \sqrt{481}}{-32} \approx 1.15$

Time must be positive, so the result is 1.15 seconds.