(h)t=-4.9t^2+98t+8 Find the max height?
are you taking Calculus ?
then
h'(t) = -9.8t + 98 = 0
t = 10
h(10) = -4.9(100) + 98(10) + 8 = 498
or
complete the square
h(t) = -4.9(t^2 - 20t + 100 - 100) + 8
= -4.9( (t-10)^2 - 100) + 8
= -4.9(T-10)^2 + 490+8
= -4.9(t-10)^2 +498
vertex is (10, 498) , so max height is 498
To find the maximum height of the equation h(t) = -4.9t^2 + 98t + 8, we can use the vertex form of a quadratic equation.
The vertex form is given by h(t) = a(t - h)^2 + k, where (h, k) represents the vertex of the parabola.
The coefficient 'a' determines whether the parabola opens upwards or downwards. In our equation, the coefficient 'a' is -4.9, which means the parabola opens downwards.
To find the vertex of the parabola, we need to find the x-coordinate of the vertex.
The x-coordinate of the vertex is given by the formula: t = -b / (2a).
From our equation, a = -4.9 and b = 98, so we can substitute these values:
t = -98 / (2 * -4.9)
t = -98 / -9.8
t = 10
So, the x-coordinate of the vertex is t = 10.
Now, we can substitute this value back into the equation to find the y-coordinate (maximum height):
h(10) = -4.9 * (10)^2 + 98 * 10 + 8
h(10) = -4.9 * 100 + 980 + 8
h(10) = -490 + 980 + 8
h(10) = 498
Therefore, the maximum height of the equation h(t) = -4.9t^2 + 98t + 8 is 498 units.