A ball is shot upward from the ground with an initial velocity of 12.0 m/s,
along a y-axis that extends upward and that has its origin y = 0 on the
ground. How much time does the ball take to move upward from y = 3.00 m
to y = 5.00 m?
To solve this problem, we need to determine the time it takes for the ball to move from y = 3.00 m to y = 5.00 m.
We can use the equations of motion to solve this problem. The equation we will use is:
y = v0t + (1/2)at^2
Where:
- y is the vertical position or displacement
- v0 is the initial velocity
- t is the time
- a is the acceleration (in this case, due to gravity)
In this case, the initial velocity is 12.0 m/s, and the acceleration due to gravity is -9.8 m/s^2 (since the ball is moving upward). The initial position is y = 3.00 m, and the final position is y = 5.00 m.
We can rearrange the equation to solve for time:
y = v0t + (1/2)at^2
Rearrange the equation to isolate t:
(1/2)at^2 + v0t - y = 0
Substituting the given values into the equation, we have:
(1/2)(-9.8)t^2 + (12.0)t - 3.00 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = (1/2)(-9.8), b = 12.0, and c = -3.00.
Plugging these values into the quadratic formula, we have:
t = (-(12.0) ± √((12.0)^2 - 4((1/2)(-9.8))(-3.00))) / 2((1/2)(-9.8))
Simplifying the equation, we have:
t = (-(12.0) ± √(144 + 58.8)) / (-9.8)
t = (-(12.0) ± √(202.8)) / (-9.8)
Taking the square root and simplifying further:
t≈(12.0 ± 14.2256) / (-9.8)
Now we can solve for both possible values of t:
t1 = (12.0 + 14.2256) / (-9.8) ≈ -2.046 s
t2 = (12.0 - 14.2256) / (-9.8) ≈ 0.224 s
Since time cannot be negative, we discard t1 and take t2 as the time it takes for the ball to move from y = 3.00 m to y = 5.00 m.
Therefore, the ball takes approximately 0.224 s to move upward from y = 3.00 m to y = 5.00 m.