If a ball is dropped from a staircase railing with a horizontal velocity of 1.09m/s which step will the ball land on? the rise of each step is 0.14m the run of each step is 0.3m. The vertical distance, from the top of the railing to the bottom of the floor is 4.8m. the slop of the steps is 0.47. my teacher says to use 4.8m-0.47(Xx) as my vertical displacement.
The slope of the steps is 0.47. Each step has a rise of 0.14 m and a run of 0.3 m. The height of the staircase is 4.8 m.
A ball is rolled with a horizontal velocity of 1.09 m/s. Let the ball land on the nth step from the top of the staircase. The vertical distance traveled by the ball is 0.14*n and the horizontal distance traveled by the ball is 0.3*n m.
When the ball is dropped there is no change in its horizontal velocity, the vertical velocity increases due to gravitational acceleration by 9.8 m/s^2.
If the ball takes time t to land on the nth step, `1.09*t = 0.3*n` and `(1/2)*9.8*t^2 = 0.14*n`
`1.09*t = 0.3*n => t = (0.3/1.09)*n`
Substitute in `(1/2)*9.8*t^2 = 0.14*n`
=> `(1/2)*9.8*((0.3/1.09)*n)^2 = 0.14*n`
=> `4.9*(0.3/1.09)^2*n = 0.14`
=> `n = (0.14/4.9)*(1.09/0.3)^2`
=> `n ~~ 0.377`
The ball falls on the next step from which it is rolled off.