Question

d. 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.
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Answer 1

Answer:

Explanation:

(i)                  When stone throw upward with initial velocity, u

Final velocity at maximum height, v=0  

Apply kinematic equation of motion.

v=u+at

t = v-u/a

= 0-u/-g

Apply kinematic equation of motion.

s = ut + 1/2 at^2

h = u(u/g) - 1/2g(u/g)^2

u=  √2gh

Time,

t = u/g = √2 gh

t = √2h/g......................(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 = v-u/a  = √2gh - 0/g   = √2h/g ............(2)

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

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Answer 2

Answer:

hopenit helps u

Explanation:

Show that the time taken to go up is same as the time taken to come down. ... when the stone reaches the maximum height after time t , its final velocity is zero. When the stone travelling vertically upwards, it is experiencing the retardation due to gravity.

and at last good night