A marker is rolled horizontally off the top of a table. after 5 econds the marker lands on the ground with a final velocity of -2.5 m/s which kinematic equation would be most useful for finding the balls initial velocity (Assume a=-9.8 m/s^2)
The appropriate kinematic equation for finding the ball's initial velocity would be:
v_f = v_i + at
Where:
- v_f is the final velocity (in this case, -2.5 m/s)
- v_i is the initial velocity (unknown)
- a is the acceleration due to gravity (-9.8 m/s^2)
- t is the time elapsed (5 seconds)
Rearranging the equation to solve for v_i:
v_i = v_f - at
Substituting the given values:
v_i = -2.5 m/s - (-9.8 m/s^2)(5 s)
v_i = -2.5 m/s + 49 m/s
v_i = 46.5 m/s
Therefore, the initial velocity of the marker is 46.5 m/s.
To find the marker's initial velocity, we can use the kinematic equation that relates time, initial velocity, final velocity, and acceleration:
v = u + at
Where:
v = final velocity (-2.5 m/s)
u = initial velocity (unknown)
a = acceleration (-9.8 m/s^2)
t = time (5 seconds)
Rearranging the equation to solve for initial velocity (u), we have:
u = v - at
Substituting the given values:
u = -2.5 m/s - (-9.8 m/s^2 * 5 s)
Simplifying the expression:
u = -2.5 m/s + 49 m/s
u = 46.5 m/s
Therefore, the initial velocity of the marker is 46.5 m/s.
To find the initial velocity of the marker, we can use the kinematic equation that relates final velocity (vf), initial velocity (vi), acceleration (a), and time (t):
vf = vi + at
In this case, the final velocity (vf) is -2.5 m/s (negative because it is directed downward) and the acceleration (a) is -9.8 m/s^2 (negative because it is directed opposite to the vertical motion). We are given the time (t) as 5 seconds.
Plugging in the given values, the kinematic equation becomes:
-2.5 m/s = vi + (-9.8 m/s^2) * 5 s
Now we can solve for the initial velocity (vi). Rearranging the equation:
vi = -2.5 m/s + 9.8 m/s^2 * 5 s
vi = -2.5 m/s + 49 m/s
vi = 46.5 m/s
Therefore, the initial velocity of the marker when rolled off the table is approximately 46.5 m/s.