Question

Guys plz answer this question

Answer 1

Answer:

First Method

Just Apply Third Equation Of Motion

v2- u2 = 2as

a= acceleration

s= displacement

u=initial velocity

v=final velocity

If the body is freely falling under gravity then this equation can be modified as

v2- u2 = 2gh

h= height from which body is dropped

g=9.8 m/s2

Now Come To Your Question According to you question body is dropped from height of 10m so we can consider its initial velicity as zero

h=10m

u=0

g=acceleration due to gravity on earth which is equal to 9.8 m/s2

Now put the values of { height, initial velocity and acceleration due to gravity in third equation of motion }

i.e

v2- u2 = 2gh

v2- 0 = 2×9.8×10

v2 = 196

v=14m/s

Hence the velocity with which it strikes the ground is 14m/s

Second Method

Now this question can also be solved by another way

mgh = 1/2×mv2

wher m is the mass of the body v is velocity and g is acceleration due to gravity and h is the height

2gh=v2

2×9.8×10 = v2

v2 = 196

v= 14m/s

By any way you solve there is no role of mass here

Explanation:

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