Question

A body is thrown vertically up with an ini
tial velocity of 40 m/s. Its velocity after 2sec
seconds will be
m/s (take
= 10m/s²)
1) 10
2) 20
3) 30
4) 40
​

Answer 1

The answer is 20 jdmdjdkskksksksoossiisis

Answer 2

Answer:

Hii mate:-

Your solution:-

Initial velocity(u) = 40m/s

Initial velocity(u) = 40m/sFinal velocity when stone reaches at max. height(v) = 0m/s

Initial velocity(u) = 40m/sFinal velocity when stone reaches at max. height(v) = 0m/sg( accln due to gravity)= 10m/s2

Initial velocity(u) = 40m/sFinal velocity when stone reaches at max. height(v) = 0m/sg( accln due to gravity)= 10m/s2g is taken as -ve because it opposes the vertical motion

Initial velocity(u) = 40m/sFinal velocity when stone reaches at max. height(v) = 0m/sg( accln due to gravity)= 10m/s2g is taken as -ve because it opposes the vertical motionmax. height of the stone reach (h) = ?

Initial velocity(u) = 40m/sFinal velocity when stone reaches at max. height(v) = 0m/sg( accln due to gravity)= 10m/s2g is taken as -ve because it opposes the vertical motionmax. height of the stone reach (h) = ?Using Equation

Initial velocity(u) = 40m/sFinal velocity when stone reaches at max. height(v) = 0m/sg( accln due to gravity)= 10m/s2g is taken as -ve because it opposes the vertical motionmax. height of the stone reach (h) = ?Using Equationv2 = u2 +2as

Initial velocity(u) = 40m/sFinal velocity when stone reaches at max. height(v) = 0m/sg( accln due to gravity)= 10m/s2g is taken as -ve because it opposes the vertical motionmax. height of the stone reach (h) = ?Using Equationv2 = u2 +2asv2 - u2 = 2gh

Initial velocity(u) = 40m/sFinal velocity when stone reaches at max. height(v) = 0m/sg( accln due to gravity)= 10m/s2g is taken as -ve because it opposes the vertical motionmax. height of the stone reach (h) = ?Using Equationv2 = u2 +2asv2 - u2 = 2gh0- u2 = -2gh

Initial velocity(u) = 40m/sFinal velocity when stone reaches at max. height(v) = 0m/sg( accln due to gravity)= 10m/s2g is taken as -ve because it opposes the vertical motionmax. height of the stone reach (h) = ?Using Equationv2 = u2 +2asv2 - u2 = 2gh0- u2 = -2ghh= u2 /2g

Initial velocity(u) = 40m/sFinal velocity when stone reaches at max. height(v) = 0m/sg( accln due to gravity)= 10m/s2g is taken as -ve because it opposes the vertical motionmax. height of the stone reach (h) = ?Using Equationv2 = u2 +2asv2 - u2 = 2gh0- u2 = -2ghh= u2 /2g= (40)2 /2*10

Initial velocity(u) = 40m/sFinal velocity when stone reaches at max. height(v) = 0m/sg( accln due to gravity)= 10m/s2g is taken as -ve because it opposes the vertical motionmax. height of the stone reach (h) = ?Using Equationv2 = u2 +2asv2 - u2 = 2gh0- u2 = -2ghh= u2 /2g= (40)2 /2*10h= 80 m

Initial velocity(u) = 40m/sFinal velocity when stone reaches at max. height(v) = 0m/sg( accln due to gravity)= 10m/s2g is taken as -ve because it opposes the vertical motionmax. height of the stone reach (h) = ?Using Equationv2 = u2 +2asv2 - u2 = 2gh0- u2 = -2ghh= u2 /2g= (40)2 /2*10h= 80 mhence, max. height attained by the stone is 80 meters....

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