Question

HELPPPPPPP!!!!!!!!!!!!!!!!
A ball is thrown from an initial height of 2 meters with an initial upward velocity of 11 m/s. The ball's height h (in meters) after t seconds is given by the following.

Answer 1

Answer:

t = .64sec and t = 1.56sec

Step-by-step explanation:

This is motion that is modeled by a parabola.  The nature of a parabola is symmtrical about some vetical line.  That means that if there is an x value where y = 3, for example, when the object is traveling upwards, it will come down after it reaches its max height and be at that same y = 3 height, just at a later time.  By this, I mean that you will have 2 times where the height is 7.  Besides that, this is a second degree polynomial, so we are expecting 2 values of t.  

We are asked to find these 2 times when the height, h, is 7.  So we will fill in a 7 for h and factor to solve for t:

$7=-5t^2+11t+2$

To factor this quadratic, we need to get everything on one side of the equals sign and set it equal to 0:

$-5t^2+11t-5=0$

Factor this however your teacher has instructed you to factor quadratics (my guess is that you're probably using the quadratic formula) to get t values of:

t = .64 sec and t = 1.56 sec