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

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Given: Initial velocity $u=40\ m/s$

Gravitational acceleration, $g=10\ m/s^2$

final velocity at maximum height $=0$

To do: To find out the maximum height reached by the stone. And also to find out the net displacement and the total distance covered by the stone.

Solution:

On using the third equation of motion

$v^2=u^2-2gh$                          [$g$ is negative as the object goes upward]

$0=(40)^2-2\times10\times h$

Or $h=\frac{(40\times40)}{20}$

So, maximum height $h=80\ m$

Total Distance covered by the stone, $s=h+h=80\ m+80\ m$

Total Distance $s=160\ m$

Total displacement $=0$             [Because the first point is the same as the last point]
Updated on 10-Oct-2022 13:22:44