A ball is dropped from rest and falls for 5 s. What are its position and velocity at that instant?

So the variables I can define are v1=0, v2= x t=5

Other than that I don't know what to do in this problem. I'm assuming position is the height so I'm looking for height and final velocity?

actually, i got 122.5 m for the position

and 49 m/s for the final velocity

height=initialheight+initiavleocity*time + 1/2 g time^2

height=initialheight-1/2 9.8 *25
yes, do velocity the same way.

How would I know what the initial height is?

Ok, then the position is 122.5 m below the initial position, and yes, on the velocity.

To determine the position and velocity of the ball at a given time, you can use equations of motion.

Since the ball is dropped from rest, its initial velocity (v1) is 0 m/s. The time taken (t) is given as 5 seconds.

To find the position of the ball (height), you can use the equation:

s = v1 * t + (1/2) * g * t^2

Where:
s is the position or height
v1 is the initial velocity
t is the time
g is the acceleration due to gravity (approximately 9.8 m/s^2)

Substituting the values, the equation becomes:

s = 0 * 5 + (1/2) * 9.8 * (5^2)

s = 0 + 4.9 * 25

s = 122.5 meters

Therefore, the position of the ball at 5 seconds is 122.5 meters.

To find the final velocity (v2), you can use the equation:

v2 = v1 + g * t

Substituting the values, the equation becomes:

v2 = 0 + 9.8 * 5

v2 = 49 m/s

Therefore, the velocity of the ball at 5 seconds is 49 m/s.