Question

1. A BALL OF MASS 0.20 KG IS THROWN VERTICALLY UPWARD WITH AN INITIAL VELOCITY OF 20 METRE/ PER SECOND CALCULATE THE MAXIMUM POTENTIAL ENERGY IT GAINS AS IT GOES UP????​

Answer 1

Answer:

K.E= 1/2 ×m×v^2

K.E must be zero.Hence P.E=40J

K.E=0.5×0.2×20×20=40JAs it has 40 J at initial point as total energy, at some point X,P.E is maximum.Total energy=P.E+K.E40=P.E+K.EAs energy is conserved, total energy at point X also must be 40 J. For P.E to be maximum K.E must be zero.Hence P.E=40JWe can come to the conclusion that it will have the maximum value at maximum height.

Answer 2

K.E=

K.E= 2

K.E= 21

K.E= 21

K.E= 21 ×m×v

K.E= 21 ×m×v 2

K.E= 21 ×m×v 2

K.E= 21 ×m×v 2 K.E=0.5×0.2×20×20=40J

K.E= 21 ×m×v 2 K.E=0.5×0.2×20×20=40JAs it has 40 J at initial point as total energy, at some point X,P.E is maximum.

K.E= 21 ×m×v 2 K.E=0.5×0.2×20×20=40JAs it has 40 J at initial point as total energy, at some point X,P.E is maximum.Total energy=P.E+K.E

K.E= 21 ×m×v 2 K.E=0.5×0.2×20×20=40JAs it has 40 J at initial point as total energy, at some point X,P.E is maximum.Total energy=P.E+K.E40=P.E+K.E

K.E= 21 ×m×v 2 K.E=0.5×0.2×20×20=40JAs it has 40 J at initial point as total energy, at some point X,P.E is maximum.Total energy=P.E+K.E40=P.E+K.EAs energy is conserved, total energy at point X also must be 40 J. For P.E to be maximum K.E must be zero.

K.E= 21 ×m×v 2 K.E=0.5×0.2×20×20=40JAs it has 40 J at initial point as total energy, at some point X,P.E is maximum.Total energy=P.E+K.E40=P.E+K.EAs energy is conserved, total energy at point X also must be 40 J. For P.E to be maximum K.E must be zero.Hence P.E=40J

K.E= 21 ×m×v 2 K.E=0.5×0.2×20×20=40JAs it has 40 J at initial point as total energy, at some point X,P.E is maximum.Total energy=P.E+K.E40=P.E+K.EAs energy is conserved, total energy at point X also must be 40 J. For P.E to be maximum K.E must be zero.Hence P.E=40JWe can come to the conclusion that it will have the maximum value at maximum height.