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Question 748367: IF I throw a bal from a height of 25 feet at a velocity of 18 feet per sec. What is the equation? How long until it hits the ground? What is the max height?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
\ =\ -16t^2\ +\ \left(v_o\sin\varphi\right)t\ +\ h_o)
Where ) is the height at time  in seconds,  is the magnitude of the initial velocity,  is the angle the initial path makes with the ground, and  is the initial height.
You didn't specify which way you threw the ball, so if straight up, then \ =\ 1) . If straight down, the \ =\ -1) . Somewhere in between -- look it up.
Once you have substituted all of your known values, set the function equal to zero -- since the height of ground is zero -- and solve the quadratic equation for .
The maximum height is 25 feet if you throw the ball at an angle with the horizontal of anything greater than or equal to  . On the other hand, if you throw the ball upward at all, the parabolic arch described by the ball will have a maximum at the vertex of the parabola. For any parabola of the form
The  -coordinate of the vertex is found by  , and that vertex is a maximum whenever the lead coefficient is less than zero. Evaluate your function at  to find the value of the maximum.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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