InterviewSolution
Saved Bookmarks
| 1. |
A stone is released from the from the top of a tower of height 19.6m calculate its final velocity just before touching the ground. |
|
Answer» Solution :According to the equation of motion under GRAVITY. `v^(2)-u^(2)=2gs` Where, u=Initial VELOCITY of the stone=o v=Final velocity of the stone s=Height of the stone =19.6m g-Acceleration DUE to gravity `=9.8ms^(-2)` Therefore `v^(-2)-o^(2)=2xx9.8xx19.6` `V^(2)=2xx9.8xx19.6=(19.6)^(2)` `V=19.6ms^(-1)` Hence, the velocity of the stone just before TOUCHING the ground is `19.6ms^(-1)`. |
|