how high does a stone rise if it is initially thrown at a rate of 15m/sec
how high should a 2kg mass be lifted the ground at a speed of 15m/sec (g=10m/sec^2)
m g h = (1/2) m v^2
10 h = .5 * 15^2
h = .05 * 225 = 11.25 meters
first question:
PE gained= KE lost
mgh=1/2 m v^2
h= 1/2 v^2/g
second question: makes no sense to me.
Physics
To calculate the maximum height reached by a stone thrown vertically, you need to consider the laws of projectile motion. However, you haven't provided important information such as the angle at which the stone was thrown and whether there are any external factors affecting its motion (like air resistance).
If the stone was thrown directly upward (at an angle of 90 degrees to the ground), then you can use a simplified formula to determine the maximum height reached.
The formula is as follows:
Maximum Height = (Initial Velocity ^ 2) / (2 * Acceleration due to Gravity)
In this case, "Initial Velocity" refers to the speed at which the stone was thrown, which is 15m/sec. And the "Acceleration due to Gravity" is approximately 9.8m/s² on Earth.
Plugging these values into the formula, you get:
Maximum Height = (15^2) / (2 * 9.8)
Maximum Height ≈ 11.47 meters
Therefore, if the stone was thrown directly upward with an initial velocity of 15m/sec, it would reach a maximum height of approximately 11.47 meters.
Please note that this estimation assumes ideal conditions without considering other factors that could affect the stone's trajectory.