Question

A flying cannonball’s height is described by formula y=−16
`t^(2)` +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.