A projectile is thrown upward so that its distance above the ground after t sec is given by h(t)=-11t²+264t. After how many seconds does it reach its maximum height?
11 t^2 - 264 t = - h
t^2 - (264/11 )t = -(1/11) h
t^2 - (24)t + 144 = -h/11 + 144
(t-12)^2 = -h/11 + 144
t = 12 at vertex
A projectile is thrown upward so that its distance (in feet) above the ground after t seconds is given by
h left parenthesis t right parenthesis equals negative 15 t squared plus 420 t.
h(t)=−15t2+420t. What is its maximum height?
To find the time at which the projectile reaches its maximum height, we need to determine the vertex of the quadratic equation that represents the height function.
The height function is given as:
h(t) = -11t² + 264t
We note that this is a quadratic equation in the form of h(t) = at² + bt + c, where a = -11, b = 264, and c = 0.
The formula for the x-coordinate of the vertex of a quadratic equation in the form of h(t) = at² + bt + c is given by:
t = -b / (2a)
Substituting the values of a and b:
t = -264 / (2*(-11))
t = -264 / (-22)
t = 12
Therefore, the projectile reaches its maximum height after 12 seconds.
To find the time at which the projectile reaches its maximum height, we need to determine the vertex of the quadratic function h(t)=-11t²+264t. The vertex of a quadratic function is given by the formula t = -b/2a, where a is the coefficient of the quadratic term and b is the coefficient of the linear term.
In this case, the coefficient of the quadratic term is -11 (a), and the coefficient of the linear term is 264 (b). Using the formula, we can calculate the time at which the projectile reaches its maximum height:
t = -b / (2a)
t = -264 / (2(-11))
t = -264 / (-22)
t = 12
Therefore, the projectile reaches its maximum height after 12 seconds.