A metal sphere of mass 200g falls from a height of 5 m. Calculate velocity with which it will strike the ground.g=10m/s-2

### Asked by khushdab | 23rd Apr, 2020, 02:17: PM

Expert Answer:

### m = 200 g = 0.2 kg
h = 5 m
a = g = 10 m/s^{2}
u = 0 ( as it is going to fall from a rest position)
v =?
By third equation of motion, we know,
v^{2} - u^{2} = 2as
As the sphere is falling from height, equation becomes,
v^{2} - u^{2} = 2gh
v^{2} - 0 = 2 × 10× 5
v^{2} = 100
Thus,
v = 10 m/s
**OR **
**There is one more way to solve this problem, **
**we have mass, height and acceleration due to gravity. **
**When the sphere is at height, potnetial energy stored in it is, **
**P.E. = mgh = 0.2 x 10 x 5 =10 J **
**When the sphere strikes the ground, **
**K.E. = 1/2 mv**^{2}
**Thus, at ground K.E. = P.E. **
**10 = (1/2) x 0.2 x v**^{2}
Solving this,
v^{2} = 20/0.2 = 100
Thus,
v = 10 m/s
Hence, the velocity with which sphere will hit the ground is 10 m/s

^{2}

^{2}- u

^{2}= 2as

^{2}- u

^{2}= 2gh

^{2}- 0 = 2 × 10× 5

^{2}= 100

**OR**

**There is one more way to solve this problem,**

**we have mass, height and acceleration due to gravity.**

**When the sphere is at height, potnetial energy stored in it is,**

**P.E. = mgh = 0.2 x 10 x 5 =10 J**

**When the sphere strikes the ground,**

**K.E. = 1/2 mv**

^{2}**Thus, at ground K.E. = P.E.**

**10 = (1/2) x 0.2 x v**

^{2}^{2}= 20/0.2 = 100

### Answered by Shiwani Sawant | 23rd Apr, 2020, 07:51: PM

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