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Question 1029055:  please help me find
The max height Please
The total time the ball is in the air
The vertex would be nice to
Found 3 solutions by Boreal, Natolino1983, solver91311:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The vertex is the maximum height.
The t value is -b/2a and that is -(64/-32) or 2 seconds.
At t=2, h(t)=-64+128+90=154 feet.
Set the equation equal to 0 to find the total time the ball is in the air.
-16t^2+64t+90=0
quadratic formula
(-1/32)(-64+/- sqrt (4096+5760); sqrt (9856)=99.277)
One root is negative, which would be the left end of the parabola. The other is positive, and that is (-1/32)*(-163.277)=5.102 seconds. Notice that at x=0, y=90, the 90 is the height at which the ball was thrown. Also, the 64 represents the velocity in feet/second that the ball was thrown. Gravity is pulling back at 32 feet/sec^2, and 64/32 or 2 seconds will be the maximum height.
Answer by Natolino1983(23)  (Show Source):
You can put this solution on YOUR website! This is quite simple.
H(t) = -16t^2 + 64t + 90 =-16(t^2-4t-45/8)= -16((t-2)^2 -77/8) = 154 -16(t-2)^2.
So Max Height = 154 (when t=2).
Total Time in air, can be obtain with H(t) =0, (t-2)^2 =154/16, with t>2, t=5,1 (s).
Vertex is (154,2)
Answer by solver91311(24713)  (Show Source):
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