# math

How fast would a ball have to be thrown upward to reach a maximum height of 49 ft? [Hint: Use the discriminant of the equation
16t2 − v0t + h = 0.

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1. This equation is derived from the following:

at maximum height the velocity is zero.
y(current height)= y0 (initial height)+v0t+.5at^2

take the derivative

32t-v0=0 so vo=32t

Plug this in for the v0 value

16t2 − (32t)t + h = 0

h=49ft

-16t^2+49=0

Solve for t:
16t^2=49
4t=7
t=7/4

Plug in t and h to solve for v0

16t2 − v0t + h = 0 where h=49 and t=7/4

16(7/4)^2-v0 (7/4) +49=0

:P

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2. 👎
2. How fast would a ball have to be thrown upward to reach a maximum height of 49 ft? [Hint: Use the discriminant of the equation 16t2 − v0t + h = 0.]

1. 👍
2. 👎

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