An alien planet has g = 2.2 m/s2. If a ball is thrown upward from a tower 4.0 m above the surface of the planet at an initial speed of 38.1 m/s, at what time or times is the ball at a height = 38.6 m? If there is more than one time, list from earliest to latest time. If there is only one time, type NA for time
To solve this problem, we can use the kinematic equations of motion. Let's consider the following variables:
s = initial height = 4.0 m
u = initial velocity = 38.1 m/s (upward)
g = acceleration due to gravity = 2.2 m/s^2 (downward)
h = final height = 38.6 m
We need to find the time(s) t when the ball is at a height of h.
The first step is to determine the time it takes for the ball to reach its maximum height. At the maximum height, the velocity of the ball becomes zero.
Using the equation:
v = u + gt
Setting v = 0, we can solve for t:
0 = 38.1 - 2.2t
2.2t = 38.1
t = 38.1 / 2.2
t ≈ 17.32 seconds
So, it takes approximately 17.32 seconds for the ball to reach its maximum height.
Next, we need to find the time it takes for the ball to fall from its maximum height to the height of 38.6 m.
Using the equation:
h = s + ut + (1/2)gt^2
Setting h = 38.6 and s = 0 (since the ball starts its descent from the maximum height), we can solve for t:
38.6 = 0 + 0 + (1/2)(2.2)t^2
38.6 = 1.1t^2
t^2 = 38.6 / 1.1
t^2 ≈ 35.09
t ≈ √35.09
t ≈ 5.92 seconds
Now, we have found two possible times:
1. The ball reaches the maximum height at t ≈ 17.32 seconds.
2. The ball falls from the maximum height to the height of 38.6 m at t ≈ 5.92 seconds.
The ball can reach the height of 38.6 m either during its ascent or its descent. Therefore, there are two times when the ball is at a height of 38.6 m:
1. During ascent: t ≈ 17.32 seconds.
2. During descent: t ≈ 5.92 seconds.
These times are listed from the earliest to the latest time.