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

a) h(t) = 50t2 – 16t + 3

b) h(t) = –16t2 + 50t + 3

c) 3 = –16t2 + 50t + h0

d) 3 = 50t2 – 16t + h0

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  b)  h(t) = –16t^2 + 50t + 3

Step-by-step explanation:

If you put the given values in the given equation, you get ...

  h(t) = at^2 + vt + h0

where a = -16, v = 50, h0 = 3

  h(t) = -16t^2 +50t +3 . . . . . matches choice B

Comment on the question

The given equation models the vertical height based on a vertical velocity of 'v'. It does not apply in the case where the horizontal velocity is v = 50 ft/s. None of the answer choices correctly models the situation where the pitcher is throwing toward the catcher.

man its nopq i just guessed and got it right so uhh

g00d lucc

step-by-step explanation: