Question

A projectile has a height given by the function ht4.9t42 153,where time,t, is in seconds and the height, h, is in meters. What is the maximum height of the function and at what time does it reach that
height?

Answer 1

Answer:

You can find the answer by using the formula t=-b/2a

Step-by-step explanation:

Remember the maximum height will be at the vertex. The x value of the vertex is your time, so use x=-b/2a except that t is your x. Then if it had also asked what the height was, you would plug that answer into your equation to find the y value of the vertex. Pretty sure your teacher is just asking you to find the x though:)

Answer 2

Using the vertex of the quadratic function, it is found that:

• The maximum height is of 243 meters.

• It takes 4.286 seconds to reach the maximum height.

The height of the projectile after t seconds is given by:

$h(t) = -4.9t^2 + 42t + 153$

Which is a quadratic function with coefficients $a = -4.9, b = 42, c = 153$.

The time it takes to hit the maximum height is the t-value of the vertex, given by:

$t_v = -\frac{b}{2a}$

Hence:

$t_v = -\frac{42}{2(-4.9)} = 4.286$

The maximum height is the output at the vertex, given by:

$h_{MAX} = -\frac{\Delta}{4a} = -\frac{b^2 - 4ac}{4a}$

Hence:

$h_{MAX} = -\frac{42^2 - 4(-4.9)(153)}{4(-4.9)} = 243$

The maximum height is of 243 meters.

A similar problem is given at https://brainly.com/question/24626341