SOLUTION: A ball is thrown upward with an initial velocity of 30 feet per second from a point that is 24 feet above the ground. The height (h) in feet of the ball at time t (in second) is gi
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Question 1166046: A ball is thrown upward with an initial velocity of 30 feet per second from a point that is 24 feet above the ground. The height (h) in feet of the ball at time t (in second) is given by the equation: h(t)= -16t^2+30t+24
a) at what time did the ball reach its maximum height?
b) what was the maximum height?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The general function for a projectile fired vertically upward near the surface of the earth at an initial velocity of 
upward and an initial height of 
with a gravitational acceleration constant 
or 
is:
\ =\ \frac{1}{2}gt^2\ +\ v_ot\ +\ h_o)
Note that the initial velocity is upward and the magnitude is represented by aa positive number while the gravitational acceleration is the result of a constant downward force and thus the magnitude is represented by a negative number.
The graph of this function is a concave down parabola. The time to reach maximum height is the value of the independent variable 
at the vertex of the parabola. This value of is given by 
, so for your problem 
. Compare this to the formula for the value of the independent variable for any parabola expressed as \ =\ ax^2\ +\ bx\ +\ c)
which is 
The maximum height is the value of the function at 
, so just evaluate )
John
My calculator said it, I believe it, that settles it
I > Ø
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