SOLUTION: The height in meters of a ball released from a ramp is given by the function h(t) = -4.9t^2 + 29.4t + 34.3, where t represents the time in seconds since the ball was released from
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Question 714629: The height in meters of a ball released from a ramp is given by the function h(t) = -4.9t^2 + 29.4t + 34.3, where t represents the time in seconds since the ball was released from the end of the ramp. Determine the time interval that the ball is above 50m.
What I have tried: -4.9t^2 + 29.4t + 34.3 +50 = 0
then I plug it into the quadratic formula but I keep getting really large decimals.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The height in meters of a ball released from a ramp is given by the function h(t) = -4.9t^2 + 29.4t + 34.3, where t represents the time in seconds since the ball was released from the end of the ramp. Determine the time interval that the ball is above 50m.
What I have tried: -4.9t^2 + 29.4t + 34.3 +50 = 0
:
-4.9t^2 + 29.4t + 34.3 = 50
so you should have
-4.9t^2 + 29.4t + 34.3 - 50 = 0
-4.9t^2 + 29.4t - 15.5 = 0
The ball will be above 50 m between the two x intercept solutions
I got:
x ~ 5.416 - .584 ~ 4.8 sec above 50 m
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