Asked by Casey
A ball is thrown into the air. The height h, in feet, of the ball can be modeled by the equation h= -16t^2+20t+6, where t is the time, in seconds, the ball is in the air. When will the ball hit the ground?
first should I use x= -b/2a?
first should I use x= -b/2a?
Answers
Answered by
Kuai
-2(8t^2 -10t -6) = 0
-2(2t -3) (4t +1) = 0
2t -3 = 0
2t-3 + 3 = 0 + 3
2t = 3
t = 3/2 second
Answered by
MathMate
Solve the equation
-16t^2+20t+6 = 0
will give you the times that the ball is on the ground. The negative root accounts for the fact that the ball is 6 feet above ground at t=0.
I get t=3/2 and t=-1/4
-16t^2+20t+6 = 0
will give you the times that the ball is on the ground. The negative root accounts for the fact that the ball is 6 feet above ground at t=0.
I get t=3/2 and t=-1/4
Answered by
Steve
x = -b/2a will give you the vertex. That will tell you how high the ball went if you plug it in and evaluate h.
Answered by
Casey
Thank you to all
Answered by
Omya
10 feet
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