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.
upward
To compute the ball's upward acceleration, we first need to determine the time it takes for the ball to reach its maximum height. We can use the kinematic equation:
v = u + at
Where:
v = final velocity (0 m/s at the maximum height since the ball momentarily stops)
u = initial velocity (13.8 m/s)
a = acceleration (which we need to find)
t = time
Since the ball momentarily stops at its maximum height, we can set v = 0:
0 = 13.8 m/s + at
Now, let's determine the time it takes for the ball to reach its maximum height. We use another kinematic equation:
s = ut + (1/2)at^2
Where:
s = displacement (0.495 m)
u = initial velocity (13.8 m/s)
a = acceleration (which we need to find)
t = time
Plugging in the values:
0.495 m = (13.8 m/s)t + (1/2)a(t^2)
Next, we can substitute the value of t from the first equation into the second equation to eliminate t:
0.495 m = (13.8 m/s)((-13.8 m/s)/a) + (1/2)a(((-13.8 m/s)/a)^2)
Simplifying, we get:
0.495 m = (-13.8^2 m^2/s^2)/a + (1/2)a(((-13.8^2 m^2/s^2)/a^2)
Expanding and rearranging, we have:
0.495 m = -191.16 m^2/s^2/a + 95.58 m^2/s^2
Rearranging again, we get:
-0.495 m - 95.58 m^2/s^2 = -191.16 m^2/s^2/a
Simplifying further:
-96.07 m^2/s^2 = -191.16 m^2/s^2/a
To find the acceleration, we can rearrange the equation:
a = (-191.16 m^2/s^2)/(-96.07 m^2/s^2)
Calculating, we find:
a ≈ 1.992 m/s^2
Therefore, the ball's upward acceleration, assuming it to be constant, is approximately 1.992 m/s^2.