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 of the object, we can use the kinematic equations of motion.
Let's break down the given information:
Initial velocity (u) = 12 m/s (upward)
Acceleration due to gravity (g) = 9.81 m/s^2 (downward)
First, we need to find the time taken for the object to reach one-fifth of its maximum height. We can do this by using the equation:
v = u + at
Where:
v = final velocity (0 m/s at the highest point, as the object momentarily comes to rest)
u = initial velocity (12 m/s)
a = acceleration due to gravity (-9.81 m/s^2)
t = time taken
Rearranging the equation, we have:
t = (v - u) / a
Substituting the given values, we have:
t = (0 - 12) / (-9.81)
Calculating the value of 't', we get:
t ≈ 1.226 seconds
Now, we can find the maximum height (h) reached by the object using the equation:
h = ut + (1/2)at^2
Given:
u = 12 m/s
t = 1.226 seconds
a = 9.81 m/s^2
Substituting the values, we have:
h = (12 * 1.226) + (0.5 * (-9.81) * (1.226)^2)
Calculating further, we get:
h ≈ 7.783 meters
Therefore, the maximum height reached by the object is approximately 7.783 meters.