A ball rolls horizontally off the edge of a tabletop that is 1.11 m high......?

Question

A ball rolls horizontally off the edge of a tabletop that is 1.11 m high. It strikes the floor 2.80 m away from the edge of the table. What is the initial (horizontal) velocity of the ball?

Answer 1

Figure it this way : since the horizontal vector will be unchanged during the fall, the time for gravity to accelerate the ball through 1.11 m is the same as the time for its constant horizontal velocity to take it 2.80 m.

Calculate how long it takes to move 1.11 m at a constant acceleration of 9.8 m/s^2 (starting from 0.0). Then divide 2.80 m by that length of time to get the starting horizontal velocity.

Answer 2

Hint: Downward acceleration ( ay ) is g, 9.81 m^2/sec, and downward initital velocity ( vy ) is zero. Forward acceleration ( ax ) is zero. Forward initial velocity ( vx ) is v. Assume drag due to air resistance is negligible.

1. Find out how much time ( t ) is required for the ball to hit the floor using ( ay ) and distance ( d = 1.11 m) .

2. Use this time to back calculate for inital velocity. Use the equation for constant velocity ( d2 - d1) = (v2 - v1 ) t.

Note for number 2: This applies to a projection onto the forward axis. Do not include the distance travelled vertically. You used this distance to calculate time. Otherwise, you will complicate your formula and you will need to then include a single equation with i and j directions.

Answer 3

preliminary vertical velocity of the ball = 0, making use of s= ut + at^2/2, the place s = 2m , u= 0 and a = g = 10 m/s 2, u can calc. t = sqrt.2/5. Now, in this time ball travells a million.40 3 m with 0 accelaratn., so, v = a million.40 3/sqrt.2/5.