A stone weighing 3 kg falls from the top of a tower 100 m high and buries itself 2 m deep in the sand .time of penetration is?

Question

M=3kg

H1=100m

H2=102m

T=?

Answer 1

Assuming the deceleration is constant while penetrating the ground, calculate the deceleration from kinematics knowing the stopping distance, and then get the time from the acceleration and the change in speed.

 

It hits the ground at speed

 

v = √(2gh) = √[2(9.8)(100)] m/s ≅ 44.3 m/s

 

From kinematics,

 

0² - v² = 2ad ⇒ 

 

a = -v²/(2d) = -(44.3)²/[2(2)] m/s² = -490.6 m/s²

 

Penetration time to stopping:

 

Δt = Δv/a = (0 - 44.3)/(-490.6) s ≅ 0.0903 s