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
when a body is thrown at an angle in a projectile motion, the vertical component of the velocity is vcos(B) ..where v is the velocity at which the body is thrown and B represents the angle at which it is thrown.Similarly horizontal component is vsin(B). these components are useful in determining the range of the projectile ,the maximum height reached,time of ascent,time of descent etc.,
Yes, But in uniform motion only.
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)
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. 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.
Yes, both are directed downward.
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.
The maximum height attained by the body can be calculated using the formula: height = (initial velocity)^2 / (2 * acceleration due to gravity). Since the velocity is reduced to half in one second, we can calculate the initial velocity using the fact that the acceleration due to gravity is -9.81 m/s^2. Then, we can plug this initial velocity into the formula to find the maximum height reached.