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
​

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= v-u/a = 0-u/-g

Apply kinematic equation of motion.

s=ut+ 1/2at^2

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

u= root 2gh

​Time,

t= u/g= root 2gh/g

(ii) When stone drop from height h

Apply kinematic equation of motion.

v^2 - u^2 =2gh

v = root 2gh

Apply kinematic equation of motion.

v=u+at

t= v-u/a

= root 2gh - 0/g

=root 2h/g

 

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

Explanation: