Time remaining:
##### the position of a ball thrown

label Calculus
account_circle Unassigned
schedule 0 Hours
account_balance_wallet \$5

the position of a ball thrown up from the top of a building 150 ft tall with an initial velocity of 20 ft/sec is given by the function s(t) = -16r^2 +20t +150. the maximum height of the ball is

Apr 8th, 2015

The height of the ball is described by a quadratic function s(t) = at^2 + bt + c, where a = -16, b = 20, and c =150. The function attains its maximum at t = -b/(2a) = 20/(2*16) = 5/8 = 0.625 s.

Then the height is s(0.625) = - 16*(5/8)^2 + 20*(5/8) + 150 = 156.25 ft above the ground.

Apr 8th, 2015

thanks

Apr 8th, 2015

you are welcome

Apr 8th, 2015

...
Apr 8th, 2015
...
Apr 8th, 2015
Aug 22nd, 2017
check_circle
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}