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.
To compute the ball's upward acceleration, we need to use the kinematic equation:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.
Given:
u = 13.8 m/s (initial velocity)
s = 0.495 m (distance)
We need to find a (acceleration).
Since the ball is thrown upward, the final velocity when it reaches the highest point is 0 m/s. Therefore, v = 0.
Using the kinematic equation, we can rearrange it to solve for a:
0 = (13.8)^2 + 2a(0.495)
Simplifying:
0 = 190.44 + 0.99a
Rearranging:
0.99a = -190.44
Dividing both sides by 0.99, we get:
a = -192.73 m/s^2
The negative sign indicates that the acceleration is in the opposite direction to the initial velocity (upward).
Therefore, the ball's upward acceleration, assuming it to be constant, is approximately -192.73 m/s^2.