Question

A ball is thrown from an initial height of 3 feet with an initial upward velocity of 28 ft/s. The ball's height h (in feet) after 1 seconds is given by the following:
h = 3 +28t-16t^2
Find all values of 1 for which the ball's height is 11 feet.
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)
1 seconds
DO
х
?
initial
height
ground

Answer 1

The value of t  at a height of 11 feet is 0.36 sec when the ball is going upwards and 1.39 seconds when the ball is coming downwards.

What is a quadratic equation?

A quadratic equation is an equation whose leading coefficient is of second degree also the equation has only one unknown while it has 3 unknown numbers.

It is written in the form of ax²+bx+c.

Given that the height of the ball at time t is given by the function,

$h(t) = 3+ 28t -16t^2$

where t is in seconds and h is in feet.

Now, the height of the ball is given to be 11 feet, therefore, substituting the value of h in the function as 11,

$h(t) = 3+ 28t -16t^2\\\\11 = 3+ 28t -16t^2\\\\-16t^2 + 28t +3 -11 = 0\\\\-16t^2 + 28t -8 = 0$

Now, solving the given quadratic equation, we will get,

$x= \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\t = \dfrac{-(28)\pm \sqrt{(28)^2-4(-16)(-8)}}{2(-16)}\\\\t= 0.36\ or\ 1.39$

Hence, the value of t  at a height of 11 feet is 0.36 sec when the ball is going upwards and 1.39 seconds when the ball is coming downwards.

Learn more about Quadratic Equations:

brainly.com/question/2263981

#SPJ1