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a ball is thrown into the air with an initial upward velocity of 48ft/s its height h in feet after t seconds is given by the function h=-16t^2 + 48t + 4
A.how many seconds will the ball reach its maximum height
B. what is the balls maximum height
C. what do you suppose would happen if the ball was thrown with less velocity would the number of seconds reach maximum height be higher or lower

Answer :

jasminesilva2907
A=-16
B=48
C=6
V=(-48/2•(-16)’ 4•(-16)•6 -48^2 /4•(-16)
=48/-32’ 384 -2304/64)
=(1.5,30)

it reaches the max height of 42ft at 1.5 seconds

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