A coin is tossed into the air with an initial speed of 8.00 m/s. In the absence of air resistance, how high does the coin go above its point of release
It rises a distance h such that the potential energy gain (m g h) equals the initial kinetic energy loss ((1/2) m v^2 ). Therefore v^2 = 2 g h
v is the initial speed. It is assumed that the coin is tossed straight up, otherwise the answer is different
Solve for h
thanks
To find the maximum height the coin reaches above its point of release, we can use the principle of conservation of energy.
The coin starts with an initial kinetic energy due to its motion, which is equal to (1/2) m v^2, where m is the mass of the coin and v is the initial velocity.
As the coin rises, it loses its initial kinetic energy but gains potential energy due to its increasing height above its point of release. The potential energy gained is given by mgh, where g is the acceleration due to gravity and h is the height.
According to the conservation of energy, the initial kinetic energy loss is equal to the potential energy gain, so we can set up an equation as follows:
(1/2) m v^2 = mgh
Now, we can solve for h:
Multiply both sides of the equation by 2 to get rid of the fraction:
m v^2 = 2mgh
Divide both sides by m:
v^2 = 2gh
Now, solve for h:
h = v^2 / (2g)
Given that the initial speed is 8.00 m/s and the acceleration due to gravity is approximately 9.8 m/s^2, we can substitute these values into the equation:
h = (8.00 m/s)^2 / (2 * 9.8 m/s^2)
h = 64.00 m^2/s^2 / 19.6 m^2/s^2
h ≈ 3.265 m
Therefore, the coin reaches a maximum height of approximately 3.265 meters above its point of release.