A ball rolls along flat ground at 7.0 m/s then encounters a hill. If the ball rolls 15 m up the hill before coming to a stop, what was the acceleration (in m/s2) experienced by the ball as it rolled up the hill?
To find the acceleration experienced by the ball as it rolled up the hill, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s, since the ball came to a stop)
u = initial velocity (7.0 m/s, as given)
a = acceleration (unknown)
s = displacement (15 m, as given)
Rearranging the equation:
0^2 = 7.0^2 + 2a(15)
Simplifying:
0 = 49 + 30a
Subtracting 49 from both sides:
-49 = 30a
Dividing both sides by 30:
-49/30 = a
Therefore, the acceleration experienced by the ball as it rolled up the hill is approximately -1.63 m/s^2.
To find the acceleration experienced by the ball as it rolls up the hill, we can use the following equation:
v_f^2 = v_i^2 + 2aΔx
Where:
- v_f is the final velocity of the ball (which is 0 m/s since the ball comes to a stop)
- v_i is the initial velocity of the ball (which is 7.0 m/s)
- a is the acceleration of the ball while rolling up the hill
- Δx is the displacement of the ball (which is 15 m)
Rearranging the equation, we have:
a = (v_f^2 - v_i^2) / (2Δx)
Now, let's substitute the known values into the equation:
a = (0^2 - 7.0^2) / (2 * 15)
Simplifying the equation:
a = (-49) / (30)
Therefore, the acceleration experienced by the ball as it rolled up the hill is approximately -1.633 m/s^2.
Note: The negative sign indicates that the acceleration is in the opposite direction of the ball's initial velocity.