# A stone weighting 0.5kg falls from a height and strikes the ground in 4 seconds. It buries deep in sand in 0.09 seconds. The depth of penetration is (g=10m/s​2)a) 100mb) 2.0mc) 1.8md) 0.9m

Eshan
askIITians Faculty 2095 Points
4 years ago
Dear student,

Velocity of stone when it reaches the ground= u+at=40m/s

The stone stops in 0.09seconds. Therefore from third equation of motion,

$\dpi{80} a=\dfrac{-u}{t}=-\dfrac{40}{0.09}m/s^2$
Therefore the distance travelled=$\dpi{80} -\dfrac{u^2}{2a}=\dfrac{40\times 0.09m}{2}=1.8m$