#
Q2. A stone weighing 3kg from the top of a tower 100 meter high and burries itself 2m deep in the sand. the time of penetration is

Q2. A stone weighing 3kg from the top of a tower 100 meter high and burries itself 2m deep in the sand. the time of penetration is

### Asked by vikasg13.hardware | 23rd Jun, 2017, 12:55: PM

Expert Answer:

###
Let us approach this interesting question with a step-by-step method:
1) The weight of the stone is a redundant value, not used in calculation.
2) At the time of falling, its initial velocity is u=0 m/s, the displacement s=100 and acceleration due to gravity g=-9.8 m/s^{2}.
3) Applying above values in the formula, v^{2}=u^{2}+2as, we get the velocity when the stone just reaches the bottom of the tower and is about to penetrate the sand.
4) For penetration of stone inside the sand, the above calculated 'v' will now be initial velocity i.e 'u' and at the end of penetration, the final velocity v=0.
5) Applying above values in formula, v^{2}=u^{2}+2as, we get the acceleration 'a' of stone penetration in sand.
6) Using the initial velocity of sand penetration 'u', final velocity at the end of penetration v=0 and above calculated 'a' in sand in the formula v=u+at, we get the time taken to penetrate the sand by 2 m.

^{2}.

^{2}=u

^{2}+2as, we get the velocity when the stone just reaches the bottom of the tower and is about to penetrate the sand.

^{2}=u

^{2}+2as, we get the acceleration 'a' of stone penetration in sand.

### Answered by Abhijeet Mishra | 23rd Jun, 2017, 03:28: PM

## Related Videos

- Derive an equation for position -time (S=ut+1/2at^2) relation for an object that travelled a distance 's' in time 't' under uniform acceleration 'a'
- a train is moving with an initial velocity of 30 metre per second the brakes are applied so as to produce a uniform acceleration of -1.5 metre per second square calculate the timeuniforma uniform acceleration
- an object is moving with a velocity 5m/s find the distance travelled in 5sec if the object is accelerating with a uniform acceleratrion of 5 m/s
- A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is
- what is nth second of motion
- A car starts from the rest and accelerates at the rate of 5m/s2. It continue accelerating for 10 s and then move at constant velocity for next 10 s. Calculate the total distance traveled by the car in 20s. (5)
- A car is coasting backwards down a hill at -3.0 m/s when the driver gets the engine started. After 2.5s, the car is moving uphill at a velocity of +4.5 m/s. What is the car's acceleration?
- Problems and solutions on kinematics, Horizontal motion
- Please solve it with method
- Derivation of the three equation s of motion

### Kindly Sign up for a personalised experience

- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions

#### Sign Up

#### Verify mobile number

Enter the OTP sent to your number

Change