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 solve this problem, we can use the kinematic equation that relates displacement, initial velocity, final velocity, acceleration, and time:
v^2 = u^2 + 2as
Where:
v is the final velocity,
u is the initial velocity,
a is the acceleration,
s is the displacement.
Let's plug the given values into the equation and solve for acceleration (a):
Given:
u = 13.8 m/s
s = 0.495 m
From the problem statement, we know that the final velocity (v) will be zero when the ball reaches its maximum height because it stops momentarily before falling back down. Therefore, we can replace v with zero:
0 = (13.8)^2 + 2a(0.495)
Now we can solve for a:
0 = 190.44 + 0.99a
Rearranging the equation:
0.99a = -190.44
Dividing both sides by 0.99:
a = -192.36 m/s^2
The negative sign indicates that the acceleration is directed upward, opposite to the direction of the gravitational force. So, the ball's upward acceleration, assuming it's constant, is approximately -192.36 m/s^2.