Here are two different methods to solve this kind of problem.
1) Use one of the formulae for constant acceleration. In this case, vf2 = vi2 + 2as, where vf is the final velocity, vi is the initial velocity (zero in this case), a is the acceleration (9.8 meters / second2), and s is the distance.
2) Do an energy calculation, as follows: Calculate the potential energy at a height of 6 meters, with the formula PE = mgh. Since we can assume that the entire potential energy gets converted to kinetic energy just before the ball hits the ground, solve for velocity, in the kinetic energy formula.
The initial velocity of a dropped ball is zero in the y (up-down) direction. After it is dropped gravity causes an acceleration, which causes the velocity to increase. F = ma, The acceleration due to gravity creates a force on the mass of the ball.
If it was thrown horizontally or dropped, and hit the ground 3.03 seconds later, then it hit the ground moving at a speed of 29.694 meters (97.42-ft) per second. If it was tossed at any angle not horizontal, and hit the ground 3.03 seconds later, we need to know the direction it was launched, in order to calculate the speed with which it hit the ground.
The ball dropped from 4m height has more kinetic energy just before it hits the ground because it has a higher velocity due to falling from a greater height. Kinetic energy is directly proportional to both mass and the square of velocity, so the ball dropped from 4m height will have more kinetic energy than the one dropped from 2m height.
To calculate the velocity of the ball just before it hits the ground, we can use the equation of motion: velocity = acceleration x time. The acceleration due to gravity is approximately 9.8 m/s^2. Given the time of 3.0 seconds, we can plug these values into the equation to find the velocity. Therefore, the velocity of the ball just before it hits the ground is 29.4 m/s.
You can calculate the kinetic energy just before hitting the ground using the formula for potential energy and kinetic energy. First, calculate the potential energy at the initial height using mgh (mass x gravity x height). Then equate this value to the kinetic energy just before hitting the ground using the formula 1/2mv^2 (0.5 x mass x velocity squared) and solve for the velocity.
mgh = 1/2 * m * v^2 v = sqrt (2 * 9.8 * 1.2) v = 4.8 m/s [down]
31 m/s
5968
it strikes the ground at a velocity of 17.9 ft/s
The ball had potential energy before it was dropped. This potential energy was due to its position above the ground.
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.
No, both balls will hit the ground at the same time, assuming they are dropped from the same height and in a vacuum. The horizontal velocity does not affect the time it takes for an object to fall vertically due to gravity.