Question

A stone is thrown vertically upward with an initial velocity of $40\;{\text{m}}/{\text{s}}$. Taking ${\text{g}} = 10\;{\text{m}}/{{\text{s}}^2}$ find the maximum height reached by the stone. What are the net displacement and the total distance covered by the stone?

Answer 1

When a ball goes vertically upwards, with the offered velocity, it initially begins to move upwards and then comes back down. To answer this question, first we need to find when the ball moves upwards and then when it falls free from its maximum height. To find the required answer, we need to apply Newton's third equation of motion. And the total distance travelled by the ball will be covered by the addition of distance when going up and down.
Formula Used: We will use the third equation of motion
${v^2} - {u^2} = 2as$
Where
$v$ is the final velocity of the stone
$u$ is the initial velocity of the stone
$a$ is the acceleration of the stone
$s$ is the displacement of the stone

Complete Step-by-Step Solution:
The following information is given to us in the question:
Initial velocity, $u = 40\;{\text{m}}/{\text{s}}$
Acceleration due to gravity, $g = 10\;{\text{m}}/{{\text{s}}^2}$
When the ball will reach its maximum height, then at that point, the final velocity will be equal to zero, i.e.,$v = 0$
Now, let us use the formula
${v^2} - {u^2} = 2as$
Here, we will substitute $a = - g = - 10m/{s^2}$. We are using the negative sign because the acceleration due to gravity is in the opposite direction to motion.
Let the distance travelled by the stone going upwards be $h$
So, our formula becomes
${v^2} - {u^2} = 2 \times ( - g) \times h$
Upon substituting the values, we get
${0^2} - {40^2} = 2 \times ( - 10) \times h$
On solving, we get
$\therefore h = 80m$
The total distance travelled by the stone going upwards and then downwards will be
$h + h = 2h$
That is
$80m + 80m = 160m$
Also, the balls went up and then it backs to its original position, the displacement becomes zero.
Hence,
The net displacement $= 0m$
The net distance covered $= 160m$

Note: The correct sign with the correct formula should be used to solve this question. Even if a negative sign is shuffled with a positive sign, then the outcome will be incorrect. So, make sure to use the correct formula with correct sign convention.