A ball rolls off a 1.0 m high table and lands on the floor, 3.0 m away from the?
[a] since the ball rolls off the table it starts with a vertical velocity of 0. So distance it falls versus time is given by: s = (1/2)gt^2 t^2 = 2*s/g t = SQRT(2*s/g) = SQRT(2*1/9.8) t = 0.452 seconds [b] the ball travels a horizontal distance of 3 m in the time it takes to hit the floor. the horizontal velocity is just distance over time so: v = d/t = 3/0.452 v = 6.637 m/s [c] The ball is falling for 0.452 under the constant acceleration of gravity. v = g*t = 9.8*(0.452) v = 4.4296 m/s [d] The horizontal velocity, ignoring air resistance, is the same as it was when the ball left the table. velocity just before the ball hits the floor = 6.637 m/s
At what horizontal distance from the shelf does the ball land if it takes 0.4s to reach the floor?
The time taken to hit the floor is used to find out how much distance is covered in horizontal axis since horizontal velocity remains constant. Horizontal distance covered = Initial horizontal speed x time taken to reach the ground = 6 m/s x 0.4 seconds = 2.4 m horizontally
A ball rolls off a shelf with a horizontal velocity of v. At what horizontal distance from the shelf does the?
Time to fall = t =sqrt.(2h/g). Distance from shelf = (vt). Distance from shelf = v (sqrt.(2h/g)) g is gravity.
An autographed baseball rolls off of a 1.2 m high desk and strikes the floor 0.72 m away from the desk. The ac?
Since we are considering downwards motion only (at no point does the baseball travel upwards) we can define the downwards direction to be positive and therefore acceleration will be 9.81 m/s. First job is to find how long it takes the baseball to hit the floor, i.e. how long does it take to fall 1.2 metres? We can find this using a kinematic equation in the vertical direction. What information have got? d = distance = 1.2 m a = acceleration due to gravity u = initial velocity = 0 (as the ball rolls off the table the vertical velocity = 0 t = time = ? So we use the equation d = ut + 1/2 at², and since u is zero, ut is zero and the equation reduces to d = 1/2 at² and this rearranges to t = sqrt(2d/a) = 0.49487 seconds. Now there are no forces acting in the horizontal direction, meaning no acceleration in the horizontal direction and thus the horizontal velocity is constant. Remember how velocity is defined, velocity = distance/ time. Horizontal velocity is therefore horizontal distance/time = 0.72 m/0.49487 = 1.45 m/s.
A ball rolls off a table with a horizontal velocity of 4 m/s.?
(a) If it takes 0.6 seconds for it to reach the floor, what is the vertical component of the ball's velocity just before it hits the floor? (Use g = 10 m/s2.) (b) What is the horizontal component of the ball's velocity just before it hits the floor?
A ball rolls over the edge of a shelf with a horizontal velocity vo m/s, east. The height of the shelf is 2.50?
A ball rolls over the edge of a shelf. The height of the shelf is 2.50 m and the horizontal range of the ball from the base of the table is 1.60 m. How long does it take for the ball to hit the floor? and the initial velocty? It seems like there's not enough information. Is there some kind of assumption I'm suppose to catch on to? Thanks!!!
A 4.00 kg ball is on a 5.00 m ledge. If it is pushed off the ledge, how much kinetic energy will it have just before hitting the ground?
Use Conservation of Energy:Ui = Initial potential energyKi = Initial Kinetic EnergyUf = Final potential energyKf = Final Kinetic energyUi + Ki = Uf + KfSince the ball is not moving initially, Ki = 0. Since the ball has no more potential energy at the bottom of the ledge, all Ui is transferred to Kf.Simplify to: Ui = KfU = mghh= 5.00m, m = 4.00kg, g= 9.81 m/s^2Ui = mgh = Kf5.00 m * 4.00 kg * 9.81 m/s^2 = 196.2 JFinal kinetic energy is 196.2 Joules.
A ball is dropped from a height of 45m. What will be the time to reach the ground?
The distance is 45m. (s)Acceleration is 9.8 m/s^2 (acceleration due to gravity). (a)We have to find the time (t)By the 3rd equation of motion,s = ut + 1/2 at^2 (here u is the initial velocity of the ball)Since the initial velocity was 0 m/s, the formula becomes s = 1/2 at^2.(substituting the values in the formula)45 = 1/2 x 9.8 x t^245 = 9.8/2 x t^245 = 4.9 x t^2 (9.8/2 = 4.9)(divide 4.9 on both sides)45/4.9 = 4.9 x t^2 / 4.99.1836 = t^23.030 = tSo it takes approximately 3 seconds for the ball to reach the surface of the earth (neglecting air resistance).
A ball rolls off the edge of a stairway with a horizontal velocity of 1.8m/s. the steps are 0.2m high and 0.2m wide. Which step will the ball hit first?
The scenario we are given is this :Now I did the calculations and found out that it would land on the 4th step. How?First of all, try to imagine the situation in your mind. You should be able to realise that the only force acting on the ball is the force of gravity (I ignored air resistance, because it would make this problem a little more complex). The force of gravity, mg, is acting vertically downwards, which gives our ball an acceleration in the downwards direction, equal to g = 9.8m/s^2.In the horizontal direction, there is no net force (again, ignore air resistance). So the horizontal component of the velocity is constant throughout the projectile motion.Now, there are two ways to solve this problem, the short method and the long method.The Long Method (Trial and Error method):Well that was the long method. It makes more sense and is easier to visualise, but took a lot of time.2. The Short Method :That was quicker, bit involved more mathematics and equations and all that.I personally prefer the 2nd Method, and then checking if I got the right answer by the long method, but only doing the end part where I put S(y) = 0.8m (the 4th step)