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.