A ball is thrown upward. It leaves the hand with a velocity of 13.8 m/s, having been accelerated through a distance of 0.495 m. Compute the ball's upward acceleration, assuming it to be constant.
Whats the upward acceleration?
vf^2=vi^2+2ad
solve for a.
To determine the ball's upward acceleration, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- s is the displacement
Given:
u = 13.8 m/s (initial velocity)
s = 0.495 m (displacement)
We need to find a (acceleration).
To solve this equation for acceleration, we rearrange the equation as follows:
a = (v^2 - u^2) / (2s)
However, we don't have the value for the final velocity (v) of the ball. We can determine it using the fact that at the highest point of the ball's trajectory, the vertical velocity becomes zero.
The vertical component of velocity (v) at the highest point can be determined using the formula:
v = u + at
Where:
t is the time it takes for the ball to reach its highest point.
Since the ball is thrown upwards and returns back to the same level, the time it takes to reach the highest point is half of the total time (t/2).
To find the total time taken (t), we can use the equation:
v = u + at
Where:
v = 0 m/s (final velocity at the highest point)
u = 13.8 m/s (initial velocity)
Rearranging the equation, we get:
t = (v - u) / a
Substituting the values, we have:
0 = 13.8 - at/2
Rearranging this equation to solve for t, we get:
t = 27.6 / a
Substituting the value of t in terms of a back into the main equation:
0 = 13.8 - a * (27.6 / a) / 2
Simplifying further:
0 = 13.8 - 27.6 / 2
Solving for a:
13.8 = 27.6 / 2
a = (27.6 / 2) / 13.8
a = 1 m/s^2
Therefore, the ball's upward acceleration is 1 m/s^2.