Question 719283: The height in feet of a ball tossed in the air is given by h=-16t^2 + 10t, where t is the time in seconds. What is the initial height of the ball? a) 16 ft b)15 ft c)29 ft or d)30 ft
Question: I am the parent trying to help out. Can this problem be solved with 2 unknowns? We got this far: -16t^2 + 10 t - h = 0
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Properly written, your function is:
\ =\ -\frac{1}{2}gt^2\ +\ v_0t\ +\ h_0)
Where ) is the height of the projectile after  seconds have elapsed since launch of the projectile,  is the initial velocity in feet per second,  is the initial height in feet, and  is the acceleration due to gravity in feet per second per second -- near the surface of the Earth,  .
Your initial velocity is evidently 10 feet per second and, since there is no constant term, we have to presume that the initial height is zero, so plugging in what we know,
\ =\ -16t^2\ +\ 10t\ + 0)
You tried
which tells us the same thing about the initial height. The intial height is the height at time zero, so what you have really written is:
Hence your initial height  must, perforce, be zero. I'm not sure how you would go about throwing a ball upward when you are releasing it at ground level -- unless you were standing in an appropriately deep hole when you threw it, I suppose.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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