- #1
VKnopp
- 12
- 2
I have decided to self-teach myself A-Level Physics. I am confused about one particular step of a solution to a certain problem that the textbook provided.
Problem:
A stone is thrown vertically upwards at ##15 ms^{-1}##. How long is the stone in the air until hitting the ground?
There are a few assumptions:
-air resistance is minimal
-displacement and velocity are positive upwards and negative downwards
-acceleration is always downward and therefore negative
-acceleration due to gravity is constant
Solution:
Time of flight is ##2t##, twice time to maximum height. Maximum height is when final velocity ##v=0##.
They use the classic equation.
##v=u+at##
##v=0##, ##u=15 ms^{-1}## and gravity is ##9.8 ms^{-2}##.
##0=15+(-9.8)t##
Solving for ##t## yields ##t=1.5306##. I understand all the way up to here.
They then multiply their solution for ##t## by a factor of ##2## to get the solution. I do not understand why they do this. Why do they do this step?
Problem:
A stone is thrown vertically upwards at ##15 ms^{-1}##. How long is the stone in the air until hitting the ground?
There are a few assumptions:
-air resistance is minimal
-displacement and velocity are positive upwards and negative downwards
-acceleration is always downward and therefore negative
-acceleration due to gravity is constant
Solution:
Time of flight is ##2t##, twice time to maximum height. Maximum height is when final velocity ##v=0##.
They use the classic equation.
##v=u+at##
##v=0##, ##u=15 ms^{-1}## and gravity is ##9.8 ms^{-2}##.
##0=15+(-9.8)t##
Solving for ##t## yields ##t=1.5306##. I understand all the way up to here.
They then multiply their solution for ##t## by a factor of ##2## to get the solution. I do not understand why they do this. Why do they do this step?