A softball pitcher throws a softball to a catcher behind home plate. The softball is 3 feet above the ground when it leaves the pitcher’s hand at a velocity of 50 feet per second. If the softball’s acceleration is –16 ft/s2, which quadratic equation models the situation correctly? h(t) = at2 + vt + h0 h(t) = 50t2 – 16t + 3 h(t) = –16t2 + 50t + 3 3 = –16t2 + 50t + h0 3 = 50t2 – 16t + h0

I think the correct answer from the choices listed above is the second option. The quadratic equation that would best model the system described above would be h(t) = –16t2 + 50t + 3. We can check it by looking at the resulting units.

The resulting units should be in meters.

m [=] m/s²(s²) + m/s (s) + m [=] m -----> equal

So, it should be the correct answer.

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MsEHolt MsEHolt · Answer 2 · 2018-08-29T17:33:01-04:00

The correct answer is:

h(t) = –16t² + 50t + 3

Explanation:

The general form of an equation such as this is h(t) = at² + v₀t + h₀, where a is the constant due to gravity, v₀ is the initial velocity and h₀ is the initial height.

We are given that the constant due to gravity is -16.

The initial velocity is 50, and the initial height is 3; this gives us the equation

h(t) = -16t² + 50t + 3