A body is thrown vertically upwards with a velocity and caught back. a) what is the displacement and distance traveled ? b) how do the displacement and distance change if its velocity of is half?
Answers
a) Displacement = 0
Distance 2 * max height it achieved
Step-by-step explanation:
Since it was thrown upwards and was caught again its total displacement is zero because it came at rest at the same place it started its journey from (I.e. starting point is the end point)
Now suppose it reached a height of x meters when it was thrown so the distance it covered was 2x meters (x meter while going up and again x meters in opposite direction.)
If the velocity is halved then the distance will still be zero and the distance will be quartered (I.e. it will be divided by 4)
This happens because of The law of conservation of energy.
Eg.
Let the mass be m kg
Velocity be v m/s
Height achieved be h
Momentum = Mass * Height * gg
(1/2) mv^2 = mgh
divide both sides by m so,
v^2 = 2gh
Or, h = (v^2)/2g
If the velocity is halved then
(1/2) m(v/2)^2 = mgh
Or, mv^2/4 = 2mgh
Again cancel out the m
v^2/4 = 2gh
h =(v^2)/8g
Answer:
Step-by-step explanation: