neglecting air resistance,
distance = 1/2 gt^2 where g = acceleration of gravity = 9./8 m/sec/sec
70 = 1/2 (9.8) t^2
14.28 = t^2
t = square root 14.28 = 3.78 seconds
If a ball were to double in speed every second for ten seconds, then it would travel 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 = 1023 feet in ten seconds. This is equal to 210 - 1 feet.
I think its the dropping of a golf ball off a building! This is because the formula for velocity when something is dropped is a quadratic formula, that is of degree 2.
No,th e potential energy is the same - weight times height of 3 feet
7.07
Rest position or equilibrium position occurs when all the forces (gravity, the wind, friction etc) exerted on an object are equal.For example, a rolling ball is not in equilibrium because one or more forces (gravity or the force you used to initially move the ball) are greater in strength than the friction (both between the ball and the surface it is rolling on and between the ball and the air).The ball will stop rolling when the force causing it to move 'forwards' is overcome by frictional force. The ball will then be in equilibrium, or at rest.
a. 144 feet b. 96 ft/sec.
The time it takes for a ball to hit the ground when dropped from a height can be calculated using the equation: t = √(2h/g), where h is the height (443 meters) and g is the acceleration due to gravity (9.81 m/s²). Solving for t gives a time of approximately 9 seconds.
Still accelerating til it hits earth. ====================================== The height from which she dropped the ball is irrelevant. In any case, the ball was most likely moving at the greatest speed just as it hit the ground. The answer to the question is: zero.
The higher the ball is dropped from, the higher it will bounce back. This is due to potential energy converting to kinetic energy upon impact with the ground, propelling the ball higher when dropped from greater heights. Ultimately, the bounce height depends on factors like gravity, air resistance, and the material of the ball.
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.
The height from which the ball is dropped is the independent variable, as it is what is being manipulated. The height of the ball's bounce is the dependent variable, as it is what is being measured and is affected by the height from which the ball is dropped.
If dropped from the same height, they will hit the ground at the same time.
Assuming both were dropped from the same height above ground, in a vacuum both would hit the ground at the same time. In a significant atmosphere (e.g. average ground-level on Earch) the bowling ball would hit the ground first.
89
Yes
381 metres
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.