Home » Ask & Discuss » Physics. » Mechanics « Back to Discussion

Mechanics

New kid on the Block

Joined:	18 Sep 2007
Post:	1

18 Sep 2007 18:35:10 IST

0 People liked this

344

Kinematics-H.C Verma-Page53-Book-1

None

Q29. A ball is dropped from a height.If it takes 0.200s to cross the last 6.0m before hitting the ground,find the height from which it was dropped.g=10m/s2.

Q30. A ball is dropped from a height of 5m onto a sandy floor and penetrates the sand up to 10cm before coming to rest.Find the retardation of the ball in the sand assuming it to be uniform.

abhishek

Share this article on:

Comments (5)

ramyani chakrabarty

Blazing goIITian

Joined: 22 Apr 2007

Posts: 2537

18 Sep 2007 18:58:46 IST

Like 0 people liked this

Q30. A ball is dropped from a height of 5m onto a sandy floor and penetrates the sand up to 10cm before coming to rest.Find the retardation of the ball in the sand assuming it to be uniform.

Soln :

       Let us first find the velocity v of the ball at the instant it touches the sand.
Initial velocity u = 0
Final velocity v = ?
Distance h = 5 m

Use   v²= u² + 2gh = >   v = ^{[ ]} (2gh)

Now at the instant it touches the sand , we again use the same fomula.
But here
                Initial velocity u = ^{[ ]} (2gh)
                Final velocity v = 0

Then, 0 = 2gh + 2a ( 10 cm)

where a = acceleration and S = 0.1m

       0 = 2*10*5 + 2a* 0.1
whence a = -- 500 m / s² ( -- ve sign means retardation )

A K

Blazing goIITian

Joined: 13 Mar 2007

Posts: 700

18 Sep 2007 19:13:56 IST

Like 0 people liked this

Let h be the height from which the ball is thrown

since the ball was dropped from h so, u=o

hence v²=2gh

where v is the velocity on reaching the ground

let velocity at height 6 m from ground be u

so,

v²=u²+2g(6)

and 6=u(0.2)+1/2g(0.04)

6=u(0.2)+0.2

5.8=0.2u

u=29m/s

2gh=29²+120

2gh=841+120

20h=961

h=961/20=48.05

ramyani chakrabarty

Blazing goIITian

Joined: 22 Apr 2007

Posts: 2537

18 Sep 2007 20:22:03 IST

Like 0 people liked this

Q29. A ball is dropped from a height.If it takes 0.200s to cross the last 6.0m before hitting the ground,find the height from which it was dropped.g=10m/s2.

Soln :

We use the formula S = ut + ( 1/2 ) at² ( symbols with usual meaning )
Here, S = 6 m
         t = 0.200
         a = g = 10 m / s²
[ please note that we take g +ve as the ball is falling .Or, displacement and acceleration are in the same direction )

6 = u * 0.2 + 5 * ( 0.2 )² = > u = 29 m / s which is the velocity at the height 6 m above the ground.
    Next, if S¹ is the height from the top to the point where its velocity become 29 m / s .

            29² = 2gS¹ = 2*10*S¹

S¹ = 42.05

Therefore, original height = 42.05 + 6 = 48.05 m

shami pratap

Cool goIITian

Joined: 5 Aug 2007

Posts: 38

18 Sep 2007 22:07:26 IST

Like 0 people liked this

solve Sqrt(2h/g)-Sqrt{2(h-6)/g}=0.2. This is tme difference in falling height h and(h-6). U will get h=42.05+6=48.05

Krishna Gopal Singh

Forum Expert

Joined: 29 Dec 2006

Posts: 5153

19 Sep 2007 22:35:17 IST

Like 0 people liked this

Abhishek one question at a time. For your first question apurv and rmayani's answer's are perfect

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team