If a baseball is projected upward from ground level with an initial velocity of 64 feet per second, then its height is a function of time, give by g(t)= -16^2 + 64t. What is the maximum heigh reached by the ball?
max height is when g is max
I think you meant
g(t)=-16t^2+64t
g(t)=16t(4-t) or g=0 at t=0 and t=4
so max height is at t=2, the mid point
maxheight= -16*4+64*2=64meters
To find the maximum height reached by the ball, we need to determine the vertex of the parabolic function g(t) = -16t^2 + 64t.
The vertex of a parabola in the form f(x) = ax^2 + bx + c can be found using the formula:
x = -b / (2a)
where x represents the x-coordinate of the vertex. In this case, x represents time (t).
For our function g(t), a = -16 and b = 64. Plugging these values into the formula, we get:
t = -64 / (2 * -16)
t = -64 / -32
t = 2
Therefore, the ball reaches its maximum height after 2 seconds.
To find the maximum height, we substitute this value of t back into the original function:
g(t) = -16t^2 + 64t
g(2) = -16(2)^2 + 64(2)
g(2) = -16(4) + 128
g(2) = -64 + 128
g(2) = 64
Hence, the maximum height reached by the ball is 64 feet.
To find the maximum height reached by the ball, we need to determine the highest point on the path which corresponds to the vertex of the parabolic function g(t) = -16t^2 + 64t.
The vertex of a parabola given by the function f(t) = at^2 + bt + c is given by the formula t = -b / (2a). In this case, a = -16 and b = 64.
Using the formula, we can find the time at which the ball reaches its maximum height:
t = -b / (2a)
t = -64 / (2*(-16))
t = -64 / (-32)
t = 2 seconds
Now, we can find the height at t = 2 seconds by substituting this value into the equation g(t) = -16t^2 + 64t:
g(2) = -16*(2)^2 + 64*(2)
g(2) = -16*4 + 64*2
g(2) = -64 + 128
g(2) = 64
Therefore, the maximum height reached by the ball is 64 feet.