A cannon ball has been catapulted through the air and follows the path created by the function f(t)= -16t^2+30t+7 where f(t) is the height of the cannonball in meters at any given time (t) in seconds.

1. What is the initial height (y-intercept) the cannon ball was launched from? I know this one is 7

2. When will the cannon ball hit the ground (x-intercept)? Round to the nearest tenth.

3. What is the cannon ball’s maximum height? Round to the nearest tenth.

Guest Mar 2, 2023

#1**0 **

This is a quadratic equation, so we can solve for all of this with a graph!

sorry i cant include an image but you can graph it yourself on desmos or some other calculator

the initial height is 7 (we see this from the equation, the c therm is the y intercept (initial height))

the cannonball will hit the ground at the positive x intercept, which is (2.1,0)

the maximum height is (0.9,21.1)

hope this helps!

Nikhil Mar 2, 2023