Question

A steel ball of mass 5kg is dropped from the top of a tower. If it reaches the ground in
2 seconds
Find
(a) the velocity with which it strike the ground.

(b) the height of the building. (c) the kinetic energy of the ball just before touching

the ground and
(d) calculate the force acted on the ground, if it is stopped by the
ground in 0.01 second.​

Answer 1

Answer:

Mass of the object : 500g=0.5 kg=m

Height,s =5 m

Acceleration due to gravity=10 m/s^2=a

Initial velocity of object(from point where it is dropped=0m/s=u

Therefore, we need to find final velocity of object, so here we go

According to motion formula,

2as=v^2-u^2

2as+u^2=v^2

(2×5×10)+(0×0)=v^2

√100=v

v=10 m/s

Now, momentum=velocity ×mass

=0.5×10=5 kg m/s

Hence the momentum of the object when it hits the floor will be 5 kg m/s.

Hope this helped ;)

Answer 2

Answer:

a) 2.5 velocity

Explanation:

I think this will be the correct answer but I am not sure of it and by the way I know only the answer of first question