Question

A stone is thrown vertically upward with an initial velocity of $40m/s.$ Taking $g=10m/s^2$, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone when the stone returns to ground?

Answer 1

Step 1: Given data

Initial velocity, $u=40m/s.$

Acceleration due to gravity, $g=-10m/s^2$

Here we will consider the upward direction to be the positive and the downward direction to be negative.

At the maximum height, the final velocity $v=0.$

Step 2:Finding maximum height, distance and displacement

Applying the equation of motion,

$$\begin{gathered} v^2=u^2+2gs \\ 0={\left(40\right)}^2-2\times 10\times s \\ s=\frac{40\times 40}{2\times 10}=80m. \end{gathered}$$

Hence, the maximum height is $80m.$

Total distance covered $=s+s=80+80=160m.$

Net displacement $=0$ [ as the stone comes to the same point from where it was thrown]

Thus, the maximum height attained by stone is 80 m. The total distance traveled and displacement of the stone after reaching the ground is 160 m and 0 respectively.