A ball is thrown from an initial height of 2 feet with an initial upward velocity of 33/fts. The ball's height h (in feet) after t seconds is given by the following. =h+2−33t16t2 Find all values of t for which the ball's height is 18 feet. Round your answer(s) to the nearest hundredth. (If there is more than one answer, use the "or" button.)

t=1.283 seconds and

0.779 seconds

Step by step Explanation:

Given: h=18 ft

The given equation is h=2+33t-16t²

Then if we substitute the value of given h, h=18 ft into the given equation we have,

18=2+33t-16t²

Then if we re- arrange we have

16t²−33t+16=0

We can see that the above quadratic equation is in standard form, with a=16, b=33 and c=16 then we can use quadratic formula in solving it which is

t= −(−33±√[(−33) ²−4×16×16)]/(2×16)

= [33±√[1089−1024]/(32)

= [33±√[65]/(32)

=1.283 or 0.779 seconds

the two real roots , of the quadratic are:

1.283 and

0.779 seconds

t= 1.283 or 0.779 seconds

Hence, the ball is at 18 feet with height 0.779seconds after it has been thrown up and,

and is at 21 feet with height 1.283 seconds after after thrown down

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl