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:

Given;

Initial velocity = u;

Distance travelled(s) = h;

Diagram below shows the situation given in the question;

Explanation:

Time to go up (t1): By Newton’s first equation of the motion

v = u + at

Where v = final velocity;

u= initial velocity

a = acceleration;

t = time taken;

According to our question;

v = 0 (because object has reach a maximum height ‘h’ before coming down and velocity at maximum height is zero);

u= u;

a = -g (because when object will be going up the acceleration due to gravity will be acting downwards to make object to fall. Hence by sign convention direction of motion and acceleration is opposite therefore a is negative);

T = t1;

Putting the values in the equation of motion we get;

.

Time of descent (t2): By Newton’s second equation of motion

Where;

s = Distance travelled;

u = initial velocity;

a = acceleration;

t = time taken;

According to our question;

s = h;

u = 0 (When the object is at maximum height its velocity is zero);

a = g (because when object will be down the acceleration due to gravity will be acting downwards to make object to fall. Hence by sign convention direction of motion and acceleration is same therefore a is positive);

T = t2;

Putting the values in the equation we get;

From newton’s third equation of the motion;

Where symbols have usual meanings as above;

When the object is descending down

u = 0 (When the object is at maximum height its velocity is zero);

v = u (Final velocity will be the same as initial velocity);

 

a = g (because when object will be down the acceleration due to gravity will be acting downwards to make object to fall. Hence by sign convention direction of motion and acceleration is same therefore a is positive);

s = ;

Putting values in the equation of motion ;

Hence

Since t1 = t2 =