A falling object accelerates at a rate of 9.8 m/s2. That means that for every second that it is falling, its velocity increases by 9.8 m/s. The higher that the object is falling from, the longer it will have to speed up, thus the higher its velocity upon impact will be. (This is assuming that it does not reach terminal velocity, the velocity at which an object can no longer accelerate because it is travelling so fast that the drag force (air resistance) is equal to the force of gravity.)
Depending on whether or not you got a running start (this affects your body's velocity) you would be falling between 40 and 45 mph.
The answer will depend on the conditions under which the body is moving and how sophisticated your calculations need to be. On (or near) the surface of the earth, if the body had velocity v0 ms-1 at time t0 seconds then in the simple model, its velocity at time t seconds will be v = v0 - 9.81*t ms-1 (approx). The vertical distance will be v0t - 9.81t2 metres. In a more sophisticated model you would include air resistance. Also if v0 is large, it is possible that the height attained is large enough to affect deceleration due to gravity.
variable velocity can be defined as a moving body whose velocity changes with time
Yes, But in uniform motion only.
Yes, the average velocity of the body can be same as the instantaneous velocity at a small time interval.The values of the average and the instantaneous velocities approach each other , as the length of time interval is decreased.
The height where from the body is dropped ie h is given as h = v2 / 2g and h = (1/2) g t2 v is the final velocity and t is time of falling g = acceleration due to gravity
Kinetic energy of a falling object can be calculated for a specific height at a specific point since a falling body accelerates which means that it's velocity is changing every moment. To calculate the kinetic energy of a falling body at a certain height, we should know the mass of the body and its velocity at that point.Then we can apply the following formula: K.E. of an object = 1/2(mv2)
Falling into water from a height of around 50 feet or higher can be fatal due to the impact on the body.
No, the magnitude of a charge does not directly affect the velocity of a body. The velocity of a body depends on the forces acting on it, such as electromagnetic forces, gravitational forces, or friction. The charge of a body affects how it interacts with electric and magnetic fields, but it does not affect its velocity directly.
... accelerates at approx 9.81 metres per second squared and experiences weightlessness. Friction with the air prevents continuous acceleration and the falling body reaches a maximum velocity called the terminal velocity.
Yes, both are directed downward.
Yes. An object is in equilibrium if the velocity is constant. A constant velocity can occur if the forces balance on the object. Consider that the gravitational force is balanced by the "air resistance force", then there is no net force and thus no acceleration. Then the velocity at which this occurs will be a constant and thus the body will be in equilibrium.
For the most part, yes; once at terminal velocity, there is no acceleration, so it has direction.
Yes, as long as gravity is the only force acting on the body.
In a freely falling body, its velocity increases due to the acceleration caused by gravity. The acceleration is constant (9.8 m/s^2 on Earth), and the body's motion is only affected by gravity, not air resistance. The body's position changes continuously as it falls towards the ground.
The graph of the motion of a body falling vertically that reaches a terminal speed would show an initial acceleration until the body reaches its terminal velocity. At this point, the graph would level off, showing constant velocity as the body falls continuously.
The velocity-time graph for a body dropped from a certain height would show an initial spike in velocity as the object accelerates due to gravity, reaching a maximum velocity when air resistance equals the force of gravity. After this, the velocity would remain constant, representing free fall with a terminal velocity. When the object hits the ground, the velocity suddenly drops to zero.