A stone (mass 5 kg) drops from a cliff of 50 m and buries into 1 m in the sand. Find average resistance shown by the sand and the time it takes to bury.

As per what the given question, the following can be said:

• Mass, M = 5 kg
• Height, h = 50 m

Read Also: Newton's Third Law of Motion

Step 1: Solution to Find

According to the question, we have to find the average resistive force that is offered by sand and the  time takes to bury.

Step 2: Solution

Considering that the velocity before hitting the ground is = v

Now, from the given equation of motion, we are already aware that,

⇒ v² = u² + 2 gh

= 0 + 2 \(\times\) 9.8 \(\times\)  50

= 980

Now, when the stone buries in the sand the following can be shown:
⇒ ​v² = u² + 2 as
0 = 980 + 2 \(\times\) a \(\times\)1
Thus, a = – 490 m/s²

Now, the force that is offered by the sand is called the average resistance. From Newton’s Second Law of Motion, we are already aware that:

Thus, after replacing the values of the given formula, we can get:

= m \(\times\) a

= – 5 \(\times\) 490

= – 2, 450 N

Note: The negative sign in the solution shows that the direction of the force is opposite.

Now, in order to calculate the time for penetration, the following equation can be use:

v = u – at

On replacing the known values, we get:

0 = 31.3 − 490t 

t = 31.3/490

= 0.064 S

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