Graphs and equations

Question

John foolishly shot an arrow nearly straight up into the air from a power bow . There was no wind and the arrow tracked way up, before falling back to earth. Let h= arrow(ft) and t= time(sec). Equation h= -32(ft/sec to the two)t² + 200(ft/sec)t + 5.75(ft). At 3.125 seconds, the arrow reaches the peak of its trajectory. How high did he shoot the arrow??

Answer 1

-16t^2 + 200t +5.75

h' = -32t +200 = 0

t = 200/32 =25/4 = 6.25 seconds to reach maximum height

max height = -16(25/4)^2 +200(25/4) + 5.75 = 630.75 feet

there's a mistake in the problem's equation

should be -16 not -32 as the coefficient of the squared term