A stone is thrown vertically upwards with a velocity of 40m/s and is caught back. Taking g=10m/s2, calculate the maximum height reached by the stone. What is the net displacement and total distance covered by the stone??IF ANYONE WILL GIVE THE ANSWER FIRST, I WILL MARK HIM OR HER AS BRAINLIEST....
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Formula for calculating maximum height is h=(u)^2/2g
So answer is =(40)^2/2×10
= 1600/20
=80m
So answer is =(40)^2/2×10
= 1600/20
=80m
ArshBains:
but what is the net displacement and total distance covered by the stone.
There is no use of it .These type of statement are given only to confuse students.
But it is written in my book..
Every book has its own answer
In exam time is short so to save it we must use direct formulas.OK
ok
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