physics grade 9

Question

a pool ball leaves a 0.70-meter high table with an initial horizontal velocity of 2.6m/s. predict the (a) time required for the distance between the table's edge and the ball's landing location

Answer 1

The key to this problem is to realize that you can treat what happens in the vertical direction completely separate from what happens in the horizontal direction. We are told that initially, (time zero) the ball leaves the table with a horizontal velocity. If it's velocity is "horizontal" then it means there is NO vertical velocity. That means you have the same vertical conditions as if the ball is just dropped vertically from 0.7 meters.

So VERTICALLY, you can use the kinematic equation x = v₀t + (1/2)at² where x = -0.7 m (it's 0,7 meters down below from the initial position), v₀ = 0 m/s (there is no vertical velocity) and a = -9.8 m/s² (the acceleration due to gravity occurs in the vertical direction and is acting downwards).

x = v₀t + (1/2)at²

-0.7 = (0)t + (1/2)(-9.8)t²

-0.7 = -4.9t²

0.14286 = t²

t = √0.14286

t = 0.38 seconds

Answer 2

Excellent answer, but may I please request that you join me in using the correct UNIT next to the actual values so the students will model this as well. Units are a significant part of problem solving as a means to check that the answer is correct. Thank you for your Service and support.