For the coin that is flipped straight up we have:
a. The velocity at the top of its trajectory (at the maximum height) is 0.
b. When it reaches a high point of 0.25 m above where it was released, the initial velocity is 2.21 m/s.
c. The time that the coin spends on the air is 0.44 s.
a. The velocity of the coin at the top of its trajectory is zero because in the maximum height the ball will stop and then will start to move down.
b. When the coin reaches a high point of 0.25 m above where it was released, the initial velocity can be calculated as follows:
[tex] v_{f}^{2} = v_{0}^{2} - 2gh [/tex]
Where:
[tex] v_{f}[/tex]: is the final velocity = 0 (at the maximum height)
[tex] v_{0}[/tex]: is the initial velocity =?
g: is the acceleration due to gravity = 9.81 m/s² (it is negative because its direction is the opposite to the direction of motion of the coin)
h: is the height = 0.25 m
Hence, the initial velocity of the coin is:
[tex] v_{0} = \sqrt{2gh} = \sqrt{2*9.81 m/s^{2}*0.25 m} = 2.21 m/s [/tex]
c. The time that the coin spends on the air is given by the sum of the rise time and the fall time:
Rise time
[tex] t_{s} = -\frac{v_{f} - v_{0}}{g} = -\frac{0 - 2.21 m/s}{9.81 m/s^{2}} = 0.22 s [/tex]
Since the ball is caught at the same height as it was released, the rise time is the same as the fall time, so the total time is:
[tex] t = 0.22 s*2 = 0.44 s [/tex]
Therefore, the coin spend 0.44 seconds in the air.
You can find another example of initial and final velocity here: https://brainly.com/question/13275688?referrer=searchResults
I hope it helps you!
