Rex flips a coin vertically upwards with an initial velocity of 0.262m/s. Determine the height to which the coin will rise above the initial height. At what velocity is the coin moving when Rex catches it.
at the top, velocity is zero.
initial KE=1/2 m vi^2=final PE=mgh
solve for h.
a. V^2 = Vo^2 + 2g*h.
V = 0, Vo = 0.262 m/s, g = -9.8 m/s^2, h = ?.
b. V = Vo = 0.262 m/s.
To determine the height reached by the coin and the velocity when it is caught, we need to consider the vertical motion of the coin.
First, let's analyze the upward motion of the coin. We can use the equations of motion to calculate the height it will reach.
The equation to calculate the height (h) reached by an object launched vertically upward is:
h = v₀² / (2g)
Where:
- h is the height reached by the coin above the initial height
- v₀ is the initial velocity of the coin (0.262 m/s in this case)
- g is the acceleration due to gravity (9.8 m/s²)
Plugging in the values, we get:
h = (0.262 m/s)² / (2 * 9.8 m/s²)
Calculating that, we find:
h ≈ 0.00342 m
Therefore, the coin will rise approximately 0.00342 meters above the initial height.
Next, let's determine the velocity of the coin when Rex catches it. Since the coin is launched vertically upward and then falls back down, its final velocity when caught will be the negative of its initial velocity.
Thus, the velocity of the coin when Rex catches it will be:
v = -0.262 m/s
Therefore, the coin will be moving downward with a velocity of -0.262 m/s when Rex catches it.