A stone is thrown vertically upwards with a velocity of 40m/s and is caught back. Taking g = 10m/s , calculate the maximum height reached by the stone , what is the net displacement and total distance covered by the stone ?
A stone is thrown vertically upwards with a velocity of 40m/s and is caught back. Taking g = 10m/s , calculate the maximum height reached by the stone , what is the net displacement and total distance covered by the stone ?
Given: Initial velocity(u) = 40m/s
Final velocity when stone reaches at max. height(v) = 0m/s
g( accln due to gravity)= 10m/s2
g is taken as -ve because it opposes the vertical motion
max. height of the stone reach (h) = ?
Using Equation
v2 = u2 +2as
v2 - u2 = 2gh
0- u2 = -2gh
h= u2 /2g
= (40)2 /2*10
h= 80 m
hence, max. height attained by the stone is 80 meters
Net displacement= upward displacement + downward displacement
= 80+(-80)
= 0 meter
Since the stone will reach up and will come down
Therefore, net distance covered = 2*80m =160 meters.
Given: Initial velocity(u) = 40m/s
Final velocity when stone reaches at max. height(v) = 0m/s
g( accln due to gravity)= 10m/s2
g is taken as -ve because it opposes the vertical motion
max. height of the stone reach (h) = ?
Using Equation
v2 = u2 +2as
v2 - u2 = 2gh
0- u2 = -2gh
h= u2 /2g
= (40)2 /2*10
h= 80 m
hence, max. height attained by the stone is 80 meters
Net displacement= upward displacement + downward displacement
= 80+(-80)
= 0 meter
Since the stone will reach up and will come down
Therefore, net distance covered = 2*80m =160 meters.
