FREEFALL POSITION FUNCTION PLUS

Bless each note where each note is the initial height on dhe. In this question were given that an object is dropped from initial height on Earth and Jupiter surface, and the height of an object after time T is given by H is equal to minus one. So we'll do the of two, Which is -32 times two.

So we'll do s earth the velocity at that point. Then we want to know the physician at that point. So it takes two seconds for it to fall 108 ft.

We'll use two because we need positive time.

FREEFALL POSITION FUNCTION PLUS

And then factor completely into 18 plus 27 Times T -2, which gives us time of negative 27/8 Or two. So here we can factor out a negative too. So and then we can get everything on one side equal to zero. So we want to set negative 16 T squared minus 22 T plus 2, 20 equal to one. So when we Fall in 108 ft, That means we're to 20-108 above ground. So we need to go back to a position function and find um what time were at that position. Then we're asked to find the velocity after following 108 ft. And don't forget our units of feet per second. So we're going to want to find the velocity function by differentiating the position function And then we'll plug in three, their friend the velocity. And then the question asked us to find the instantaneous velocity at three seconds. So we can form our position function with the initial Position of 220 and Initial Velocity of -22. So um and it's also worth noting that the derivative of the position function exists, the velocity folks. Um That requires us to use the equation for an object in free fall as we'll see here as we're given here where the sub zero represents the initial velocity and as sub zero represents the initial position. is a position velocity acceleration problem.

Freefall position function