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A ball is thrown into the air with an initial upward velocity of 48 ft/s. It height h in feet after t seconds is given by the function h (t) = – 16t2 + 48t + 4.

 

A) What height will the ball be when 2 seconds has passed?

 Feb 12, 2019
 #1
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t = time    Just put   t = 2 in to the equation and calculate  h

 

h = - 16(2^2) + 48(2) + 4         Got it ?cheeky

 Feb 12, 2019
 #2
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Thank you for the help, I don't know why I missed something as simple as that.

Guest Feb 12, 2019

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