Question

A flying cannonball’s height is described by formula y=−16 title=" t^{2} " alt=" t^{2} " align="absmiddle" class="latex-formula"> +300t. Find the highest point of its trajectory. In how many seconds after the shot will cannonball be at the highest point?

Answer 1

Y = -t ^ 2 + 300t - 16
 We find the first derivative and calculate its roots.
 We make the second derivative, and calculate the sign taken in it by the roots of the first derivative, and if:
 f '' (a) <0 is a relative maximum
 f '' (a)> 0 is a relative minimum

 y '= - 2t + 300 = 0
 -2t + 300 = 0
 t = 300/2 = 150

 y'' = - 2
 y'' (150) = - 2 (is a relative maximum)

 the highest point of the trajectory is reached for t = 150s.
 The height for that time is
 y = - (150) ^ 2 + 300 (150) - 16 = 67484
 
 answer
 67484
 t = 150s.