Question

A stone thrown vertically upwards with
initial velocity u reaches a height ‘h?
before coming down. Show that the
time taken to go up is same as the time
taken to come down.
pls ans the following​

Answer 1

Answer:

When stone throw upward with initial velocity, u

Final velocity at maximum height, v=0

Apply kinematic equation of motion.

v=u+at

t=

a

v−u

=

−g

0−u

Apply kinematic equation of motion.

s=ut+

2

1

at

2

h=u(

g

u

)−

2

1

g(

g

u

)

2

u=

2gh

Time,

t=

g

u

=

g

2gh

t=

g

2h

........(1)

(ii) When stone drop from height h

Apply kinematic equation of motion.

v

2

−u

2

=2gh

v=

2gh

Apply kinematic equation of motion.

v=u+at

t=

a

v−u

=

g

2gh

−0

=

g

2h

.......(2)

From (1 ) and (2) it is proved that the time taken to go up is same as the time taken to come down.

Explanation:

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Answer 2

Answer:

When stone throw upward with initial velocity, u

Final velocity at maximum height, v=0

Apply kinematic equation of motion.

v=u+at

t= av−u

= −g0−u

Apply kinematic equation of motion.

s=ut+ 2

1

at

2

h=u( g )−

2

1

g(

g

u

)

2

u=

2gh

Time,

t=

g

u

=

g

2gh

t=

g

2h

........(1)

(ii) When stone drop from height h

Apply kinematic equation of motion.

v

2

−u

2

=2gh

v=

2gh

Apply kinematic equation of motion.

v=u+at

t=

a

v−u

=

g

2gh

−0

=

g

2h

.......(2)

From (1 ) and (2) it is proved that the time taken to go up is same as the time taken to come down.