+253 If a ball is thrown directly upward with a velocity of 26 ft/s, its height (in feet) after t seconds is given by y = 26t − 16t2. What is the maximum height attained by the ball? (Round your answer to the nearest whole number.)
+128460 y = 26t − 16t2
Rewrite as
y = -16t^2 + 26t
This is similar to the one I just did sally1, so we have
-b/2a = -26/[2(-16)] = 26/32 = 13/16 So to find max height, substitute this back into the original equation for "x" This gives....
y = -16(13/16)^2 + 26(13/6) = 169/16 = 10.5625 ft.
+128460 y = 26t − 16t2
Rewrite as
y = -16t^2 + 26t
This is similar to the one I just did sally1, so we have
-b/2a = -26/[2(-16)] = 26/32 = 13/16 So to find max height, substitute this back into the original equation for "x" This gives....
y = -16(13/16)^2 + 26(13/6) = 169/16 = 10.5625 ft.
