You have three balls, one is thrown upward at 30m/s, on is dropped, and one is thrown toward the ground at 20m/s. what is the acceleration of the dropped ball?
Vf=0m/s + (-10m/s^2)(1)
=0m/s -10m?s^2
= -10m/s
a= -10m/s * 0 /1
= 0
am i on the right track.....?
please help
a = g = 9.8 m/s^2.
You are on the right track in terms of calculating the final velocity (Vf) and time (t) when the ball is dropped. However, the formula you used to compute the acceleration is not correct.
To find the acceleration of the dropped ball, we can use the formula:
Vf = Vi + at
where:
- Vf is the final velocity,
- Vi is the initial velocity,
- a is the acceleration,
- t is the time.
In this case, the final velocity is 0 m/s since the ball stops when it reaches the ground. The initial velocity (Vi) is not given explicitly, but we can assume it is 0 m/s since the ball is dropped from rest. The time (t) is 1 second, as mentioned in your calculation.
So, using the formula above:
0 m/s = 0 m/s + a * 1 s
Now, we can solve for the acceleration (a):
0 = a
Therefore, the acceleration of the dropped ball is 0 m/s². This means that the ball does not experience any acceleration in the vertical direction when it is dropped. It simply falls due to the force of gravity without any other external forces acting on it.