Question

A ball of mass 200 g is thrown vertically with a velocity of 16 m/s. What is the P.E at the

maximum height? What is the P.E of the ball when it reaches 3/4 of the maximum height in the

upward motion?​

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.

Answer 2

Answer:

Given :-

Time = 4 sec

G = 10 m/s

u = 0 m/s

To Find :-

Height

Final velocity

Solution :-

We know that

v = u + at

=> v = 0 + (10)(4)

=> v = 0 + 40

=> v = 40 m/s

Now

=> s = (0)(4) + (1/2)(10)(4)(4)

=> s = 0 + (1/2)(80)

=> s = 0 + 40

=> s = 40 m/s