A ball of iron, mass 1 kg, is dropped from the top of the building. The ball reaches the
ground in 5s. What is the velocity, in m/s, of the ball when it strikes the ground?
A)150 ms-1
B)99 ms-1
C)49 ms-1
D]27 ms-1


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Explicație:

A ball of mass 8 kg is dropped from a height of 10 m. What is the velocity with which it strikes the ground?

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First Method

Just Apply Third Equation Of Motion

v 2 - u 2 = 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

v 2 - u 2 = 2gh

h= height from which body is dropped

g=9.8 m/s 2

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/s 2

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

i.e

v 2 - u 2 = 2gh

v 2 - 0 = 2×9.8×10

v 2 = 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×mv 2

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

2gh=v 2

2×9.8×10 = v 2

v 2 = 196

v= 14m/s

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