Question

14. An object of mass 0.25 kg is thrown vertically upwards
with an initial velocity of 40 ms. What is its initial
kinetic energy? Find the potential energy of the object
when it reaches the highest point. What is the maximum
height attained by the object? (Take g = 10 m/s)​

Answer 1

Answer:

K.E=

2

1

×m×v

2

K.E=0.5×0.2×20×20=40J

As it has 40 J at initial point as total energy, at some point X,P.E is maximum.

Total energy=P.E+K.E

40=P.E+K.E

As 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

We can come to the conclusion that it will have the maximum value at maximum height.

Explanation:

thats your answer

Answer 2

Explanation:

Kinetic energy = 1/2mv²

=1/2×1/4×40×40

=200J

according to law of conservation of energy the initial kinetic energy is equal to the potential energy at maximum height show potential energy is equal to 200 J

Hmax = u²/2g

=40×40/20

=800m