Question

S(t)=16t^2 +32t+128
How high does the ball go?
How long does it take?
When will hit the ground

Answer 1

at its max height v=0 so put ds/dt=0 and get the value of t and put it into s(t)

Answer 2

I don't understand pls walk me through

Answer 3

16t^2 has a positive coefficient so the parabola will open upwards, are you sure it's not negative 16t^2?

Answer 4

Yea it's not

Answer 5

That is not the equation of the path of an object thrown into the air.

We can solve many things for S(t)=16t^2 +32t+128
but "when the ball hits the ground" is not one of them without more info.

Answer 6

Ok soo .......

Answer 7

16x^2 draws a line that is open like a cup
-16x^2 opens down like an upside down U so that is required for the path of an object through the air.

Answer 8

Ok

Answer 9

to solve for when it hits the ground, place 0 in place of x and solve, there will be two answers, the starting point and then end point. This equation does not touch the ground you need to be sure it is correct

Answer 10

to solve for vertex (max height of travel) use -b/2(a)

Answer 11

Please recheck to see if it has a negative in front of the 16

Answer 12

There is no negative sign

Answer 13

take the first derivative to find the velocity function

Answer 14

if you plot this it looks like a U with the lowest point of travel over 100 feet off the ground. Your ball will never hit the ground