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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?
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]
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