Question

A stone is thrown vertically upward with an initial velocity of $40m{s}^{-1}$. Taking $g=10m{s}^{-2}$, the maximum displacement and distance covered by the stone on reaching the ground is

• A. Displacement = zero; distance = 80 m
• B. Displacement = zero; distance covered = 160 m
• C. Displacement = 80 m; distance = 160 m
• D. Displacement and distance = 180 m

Answer 1

The correct option is B Displacement = zero; distance covered = 160 m

Answer is B.
Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.
Acceleration due to gravity, g = $10m/s^2$ (downward motion).
Maximum height, s = H.
As the body is thrown upward a = -g the relation $v^2=u^2-2as$ gives $v^2=u^2-2aH$, we have,
H = $\frac{u^2-v^2}{2g}=\frac{{40m/s}^2-0^2}{2\left(10m/s^2\right)}=\frac{1600}{20}=80m$.
If a stone is thrown vertically upward, it returns to its initial position after achieving maximum height.
so, the net displacement = Difference of positions between initial and final positions = 0.
Total distance covered = 80 m + 80 m = 160 m.
Hence, the displacement is 0 and the total distance covered is 160 m.