you throw a ball up in the air. given the function
h=-16t^2+20t+6
1)when does the ball reach it maximum height? and
2)what is the ball's maximum height?
your equation
h= -16t^2+20t+6
is modelled by a parabola which opens downwards.
Find the vertex of this parabola by the means that you learned, and you will have both answers.
I assume you are not a calculus student.
This is a parabola. Looking at the zeroes
0=-16t^2+20t+6
( 8t+2)(-2t+3)=0
t=-1/4, t=3/2
because the maximum will occure at the halfway point between two zeroes, then
half way: (1.5+.25)/2 - 1.5
=1.5-(1.75)/2=
t=.625 is the location of the max
for max height, plug that t into the original equation.
To find the maximum height of the ball, we need to determine the vertex of the function h(t) = -16t^2 + 20t + 6.
1) To determine when the ball reaches its maximum height, we first need to find the time, t, at which the vertex occurs. The formula for finding the time of the vertex is t = -b / (2a), where a, b, and c are the coefficients of the quadratic function.
In this case, a = -16 and b = 20. Plugging these values into the formula, we get:
t = -20 / (2(-16))
t = -20 / (-32)
t = 5/8
Therefore, the ball reaches its maximum height at t = 5/8.
2) To find the maximum height, we substitute the value of t back into the function h(t). Plugging in t = 5/8, we get:
h = -16(5/8)^2 + 20(5/8) + 6
h = -16(25/64) + 100/8 + 6
h = -400/64 + 800/64 + 6
h = 400/64 + 800/64 + 6
h = (400 + 800 + 384) / 64
h = 1584 / 64
h = 24.75
Therefore, the ball's maximum height is 24.75 units.
To find the answers to your questions, we can use the given function h(t) = -16t^2 + 20t + 6, where h represents the height of the ball and t represents the time in seconds.
1) When does the ball reach its maximum height?
To find when the ball reaches its maximum height, we need to determine the value of t at the vertex of the parabolic function. The vertex of a parabola with the equation f(x) = ax^2 + bx + c is given by the formula:
t = -b / (2a)
In this case, a = -16 and b = 20. Plugging in these values, we can find the value of t:
t = -20 / (2 * -16)
t = -20 / -32
t = 0.625 seconds
So, the ball reaches its maximum height after 0.625 seconds.
2) What is the ball's maximum height?
To find the maximum height, we substitute the value of t we found for the maximum height into the function h(t):
h(0.625) = -16(0.625)^2 + 20(0.625) + 6
Evaluate the expression to find the maximum height:
h(0.625) = -16(0.390625) + 12.5 + 6
h(0.625) = -6.25 + 12.5 + 6
h(0.625) = 12.5 + 6 - 6.25
h(0.625) = 12.5
Therefore, the ball reaches a maximum height of 12.5 units.