Time of baseball flight: s = -16t^2 + v0t + s0

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If you throw a baseball straight up at a velocity of 32 feet per second. A function can be created by expressing distance above the ground, s, as a function of time, t. This function is s = -16t² + v₀t + s₀

How high is it after 1 second, what's the max height, how long before it hits the ground?

I know that's a lot. Thanks if you can help

 

[stapel]

1) To find the height at t = 1, plug "1" in for "t".

2) To find the maximum value, find the vertex.

3) To find when h = 0, plug "0" in for "h" and solve.

Eliz.

 

[stapel]

See also the reply posted here.

 

[TchrWill]

Re: Time of baseball flight

If you throw a baseball straight up at a velocity of 32 feet per second. A function can be created by expressing distance above the ground, s, as a function of time, t. This function is s = -16t2 + v0t + s0

How high is it after 1 second, what's the max height, how long before it hits the ground?

I know that's a lot. Thanks if you can help

Vo = initial velocity, Vf = final velocity, g = acceleration of gravity and t = time
Vf = Vo - 32t
Since the final velocity is 0, we have 0 = 32 - 32t making t = 1 second to zero velocity and maximum height.
Height is defined by h = Vot - 16t^2 = 32(1) - 16(1) = 16 feet.
Falling back down, Vf = 0 + 32t making t = 1 second.

In summary, the ball is 16 feet above the ground after 1 second which happens to be its maximum height. It takes 1 second to rise to its maximum height and 1 second to fall back down making the elapsed time before hitting the ground 2 seconds.