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Need help understanding the formula

A juggler is performing her act using several balls. She throws the balls up at an initial height of 4 feet, with a speed of 15 feet per second. If the juggler doesn't catch one of the balls, about how long does it take the ball to hit the floor?

Hint: Use H(t) = _16t2 + vt + s.

7.52 seconds
1.15 seconds
0.47 second
0.22 second

Respuesta :

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

[tex]h(t) = -16t^2+15t+4[/tex]

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

[tex]-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[/tex]

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

For this case we have the following equation:

[tex] H (t) = -16t ^ 2 + vt + s
[/tex]

Where,

t: time

v: initial speed

s: initial height

Substituting values we have:

[tex] H (t) = -16t ^ 2 + 15t + 4
[/tex]

By the time the ball hits the ground we have:

[tex] -16t ^ 2 + 15t + 4 = 0
[/tex]

From here, we clear the value of time:

Using the quadratic equation we have:

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

Doing the calculations we have:

[tex] t1=-0.217

t2=1.154 [/tex]

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

[tex] t=1.154 [/tex]

Answer:

it takes the ball to hit the floor about:

1.15 seconds



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