Question

A ball is thrown from the top of the tower with the velocity of 19.6 m/s vertically upward it return to earth after 6 sec. calculate the hight of the tower?​

Answer 1

Answer:

Given that,

Initial velocity u=19.6m/s

Final velocity v=0

Time t=6sec

The acceleration is

We know that,

  v=u+at

 0=19.6+a×6

 a=−3.26m/s2

Now, the height is

From equation of motion

  s=ut+21at2

 s=19.6×6−21×3.26×36

 s=58.9

 s=59m

Hence, the height of the tower is 59 m

hope it helps....

Answer 2

Given that,

Initial velocity u=19.6m/s

Final velocity v=0

Time t=6sec

The acceleration is

We know that,

v=u+at

0=19.6+a×6

a=−3.26m/s

2

Now, the height is

From equation of motion

s=ut+

2

1

at

2

s=19.6×6−

2

1

×3.26×36

s=58.9

s=59m

Hence, the height of the tower is 59 m

The work–energy theorem: -

The work done by all forces on an object is equal to the change in kinetic energy of the object.

W=Δk

From newton second law,

F=ma....(I)

We know that,

a=

dt

dv

So, put the value of a in equation (I)

F=m

dt

dv

If multiple both side by v

F.v=mv

dt

dv

.....(II)

We know that,

v=

dt

dx

Now, put the value of v in equation (II)

F

dt

dx

=mv

dt

dv

Now, on integrating

Fdx=mvdv

F

0

∫

x

dx=m

v

1

∫

v

2

vdv

Fx=

2

1

m(v

2

2

−v

1

2

)

Fx=

2

1

mv

2

2

−

2

1

mv

1

2

We know that,

W=Fx

2

1

mv

2

=k

k = kinetic energy

Now,

W=k

2

−k

1

W=Δk

The work –energy theorem is proved.

hi oppa how r u

Good night

borahae