An object is thrown vertically upward and has a speed of 12 m/s when it reaches one fifth of its maximum height above the launch point. Determine its maximum height. g=9.81 m/s^2. Please help
To determine the maximum height reached by the object, we can use the equations of motion.
Let's break down the problem step by step:
1. First, we can find the initial velocity of the object when it reaches one fifth of its maximum height (h/5) using the equation:
Final velocity (vf) = Initial velocity (vi) + Acceleration (a) * Time (t)
When the object is at one fifth of its maximum height, it is momentarily at rest before falling back down. Hence, the final velocity is zero (vf = 0).
vi = 12 m/s (given)
a = -9.81 m/s^2 (since it's opposing the object's upward motion)
t = ?
2. Rearranging the equation, we have:
0 = 12 m/s - 9.81 m/s^2 * t
3. Now, we can solve for time (t):
9.81 m/s^2 * t = 12 m/s
t = 12 m/s / 9.81 m/s^2
t ≈ 1.224 seconds
4. With the value of time, we can determine the maximum height (h) using the equation:
h = Initial velocity (vi) * Time (t) + 0.5 * Acceleration (a) * Time squared (t^2)
vi = 12 m/s (given)
t ≈ 1.224 seconds (from the previous step)
a = -9.81 m/s^2 (opposing the object's upward motion)
h = ?
5. Substituting the values into the equation, we have:
h = 12 m/s * 1.224 seconds + 0.5 * (-9.81 m/s^2) * (1.224 seconds)^2
Evaluating this expression will give you the maximum height reached by the object.
After performing the calculations, the maximum height can be found to be approximately 8.0 meters.