Question

 A flying cannonball’s height is described by formula y=−16t2+300t. Find the highest point of its trajectory. In how many seconds after the shot will cannonball be at the highest point?

Answer 1

the trajectory is 1.5497476 NE It will be at its highest point approximately 7.89 seconds after being shot Hope this helps :)

Answer 2

Answer:

The cannonball will be at the highest point in 9.375 seconds after the shot.

Step-by-step explanation:

Suppose we have a quadratic function in the format:

$y(t) = at^{2} + bt + c$

The maximum point will happen when:

$t_{v} = -\frac{b}{2a}$

The point will be:

$y_{MAX} = y(t_{v})$

In this problem:

$y = -16t^{2} + 300t$

So $a = -16, b = 300$

We have to find $t_{v}$

$t_{v} = -\frac{300}{2(-16)} = 9.375$

The cannonball will be at the highest point in 9.375 seconds after the shot.