Asked by Hey Joe
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 + 64
6 seconds
4 seconds
7 seconds
5 seconds
6 seconds
4 seconds
7 seconds
5 seconds
Answers
Answered by
oobleck
all or any of the above, depending on the question ...
Answered by
Hey Joe
Sorry, I forgot to put part of the question.
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 + 64 After how many seconds will the ball hit the ground?
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 + 64 After how many seconds will the ball hit the ground?
Answered by
oobleck
when it hits the ground, the height is zero, right?
so just solve
-16t^2 + 48t + 64 = 0
or, dividing by -16,
t^2 - 3t - 4 = 0
sure looks like t=4
so just solve
-16t^2 + 48t + 64 = 0
or, dividing by -16,
t^2 - 3t - 4 = 0
sure looks like t=4
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