- #1
Minorail
- 14
- 0
can u help simulate a ball from high h fall down
s=1/2gt^2
v=2gh
no friction in air
thank u
s=1/2gt^2
v=2gh
no friction in air
thank u
I assume you mean [itex]v = \sqrt{2gh}[/itex].Minorail said:v=2gh
The initial height (h) of the ball refers to the starting point of the ball's motion. It is the height at which the ball is released or dropped from. This height is usually measured in meters or feet.
The time it takes for the ball to drop from h is affected by several factors, including the gravitational acceleration, the initial velocity of the ball, and the air resistance. These factors can impact the speed at which the ball falls, therefore affecting the time it takes to reach the ground.
As the ball falls, its height (h) decreases. This is because the ball is moving closer to the ground, reducing the distance between the ball and the ground. The height (h) of the ball changes continuously until it reaches the ground.
The equation for calculating the time it takes for a ball to drop from h is t = √(2h/g), where t is the time in seconds, h is the initial height in meters, and g is the gravitational acceleration (9.8 m/s² on Earth).
Yes, the initial height (h) of the ball can affect its final velocity. The higher the initial height, the longer the ball has to accelerate due to the force of gravity, resulting in a higher final velocity. This is because the ball has more potential energy at higher heights, which is converted to kinetic energy as it falls.