Question 15
A stone is thrown vertically upward with an initial velocity of 40 m/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?

According to the third equation of motion under gravity:
$v^2-u^2=2gh$
Where,
u = Initial velocity of the stone
v = Final velocity of the stone
h = Displacement of the stone
g = Acceleration due to gravity

We know that velocity at the highest point is equal to zero.
Here, u = 40 m/s, v = 0, g = $=-10m{s}^{-2}$

Therefore,
$0-(40{)}^2=2\times (-10)\times h$
$h=40\times \frac{40}{20}=80m$
Maximum height reached = 80 m

Therefore, total distance covered by the stone during its upward and downward journey = 80 m + 80 m = 160 m

Net displacement of the stone during its upward and downward journey
= 80 m + (- 80 m) = 0