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 the same as the time taken to come down.

Answer 1

To find the time taken for the stone in both going up and coming down we find them separately first we find the velocity with which it travels and the time taken to travel that height, to find the time in terms of acceleration and height we use the formulas as:
For going up the velocity is given by the formula:
$u=\dfrac{S}{t}-\dfrac{1}{2}at$
For going down the velocity is given by the formula:
${{v}^{2}}\text{ }-\text{ }{{u}^{2}}\text{ }=\text{ }2gh$
The time taken for the stone to travel both distances is:
$t=\dfrac{v-u}{a}$
where u is the initial velocity, v is the final velocity, a is the acceleration, S is the distance travelled.

Complete step by step solution:
For the situation when stone goes upward.
The final velocity at the maximum height is zero as the stone stops momentarily while at the top. When the stone thrown upwards the velocity with which it goes up is
$v=u+at$
With v=0, the initial velocity is:
$0=u+at$
$u=-at$
Hence, the time taken to reach the top is:
$t=\dfrac{u}{-a}$
Now using the kinematic theorem of motion in one-dimension, we get the time for the stone to reach height "h" is:
$S=ut+\dfrac{1}{2}a{{t}^{2}}$
Placing the values of $S=h$, $t = -\dfrac{u}{a}$ and with $a=-g$ as the stone is going upwards against the gravity. After getting the time taken to reach height "h", we find the initial velocity in terms of height and gravity which is given by the formula of
$h=u\left( -\dfrac{u}{a} \right)+\dfrac{1}{2}a{{\left( -\dfrac{u}{a} \right)}^{2}}$
$h=-\dfrac{{{u}^{2}}}{a}+\dfrac{1}{2}{{\left( \dfrac{u}{a} \right)}^{2}}$
$u=\sqrt{2gh}$
Now replacing the initial velocity with the above value in the time formula, we get the time taken as:
$t=\dfrac{u}{-a}$
Acceleration is taken as gravity which is in mod.
$t=\dfrac{\sqrt{2gh}}{g}$
$t=\dfrac{\sqrt{2h}}{g}$

For the situation when stone goes down.
The velocity with which the stone goes down i.e. the initial velocity during the falling part is zero, hence, the final velocity when reaching the ground is:
${{v}^{2}}\text{ }-\text{ }{{u}^{2}}\text{ }=\text{ }2gh$
With $u=0$, the value of v is:
$v\text{ }=\text{ }\sqrt{2gh}$
Now to find the time to reach from top to bottom is given as:
$t\text{ }=\text{ }\dfrac{\left( v\text{ }-\text{ }u \right)}{a}$ derived from $v=u+at$
Now replacing the initial velocity and final velocity with the above value, we get the time taken as:
$t\text{ }=\text{ }\dfrac{\left( \sqrt{2gh}\text{ }-\text{ }0 \right)}{a}$
Acceleration is taken as gravity which is in mod.
$t\text{ }=\text{ }\sqrt{\dfrac{2h}{g}}$

Therefore, from the two time periods, going up and down is the same as $t\text{ }=\text{ }\sqrt{\dfrac{2h}{g}}$.

Note: The stone when going up and down has both initial and final velocities and the point at top when the stone is suspended in the air for a minute period of time, the velocities for both up and down are zero. The distance is the same for both up and down and for velocity in the second part that is when the stone is coming down we see that the formula for velocity is ${{v}^{2}}\text{ }-\text{ }{{u}^{2}}\text{ }=\text{ }2gh$, as the stone’s acceleration has become uniform due to free-fall which is not observed in the first part as in the first part or when stone going up the acceleration of the stone going up is against the earth’s gravitational acceleration.