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Question box-1

A boy tosses up a ball in the air with an initial velocity U .At the instance it reaches the maximum height H ,he tosses up a second ball with the same initial velocity U.Find out the height where the two balls collide?

a]H/4.

b] 3H/4

c]H/2

d] None of these
Explanation of the answer is also expecting ,Thank you.

കരിപ്പാറ സുനില്‍ said...

Namaskaram pyarilal sir , I think the falling path of the first body is different .So the chance of collition is zero

pyarilal said...

No sunil sir, Sorry I think question is not clear here, Assume two balls are in same line as one goes up and another comes down and then collide each other at a particular height. Then what may be that height?

CK Biju Paravur said...

I think the answer is H/4.
When the 1st ball is at height H. The 2nd ball starts its journey upwards. 1st ball has accelaration and the 2nd ball has retardation. So the colliding chance will be at the height H/4.

കരിപ്പാറ സുനില്‍ said...

എന്തായാലും ഉത്തരം പറയൂ പ്യാരീലാല്‍ സാര്‍ , ഒപ്പം വിശദീകരണവും .പലരും കമന്റ് ഇടുന്നില്ല എന്നേ ഉള്ളൂ . ഉത്തരത്തിനായി കാത്തിരിക്കുന്നുണ്ട് . ആശംസകളോടെ

CK Biju Paravur said...

പ്യാരിലാല്‍ സാര്‍,
ശാസ്ത്രമേളയുടെ തിരക്കൊക്കെ കഴിഞ്ഞില്ലേ....
ഇനി ഉത്തരം പറയൂ.....

pyarilal said...

answer,[b] 3H/4. now it's easy to find out How?

pyarilal said...

Ok ,I will try to explain it.To make it simple, explanation is bit lengthy. After reading it you can suggest easy and alternate explanation.
Let us consider the ball tosses up an initial velocity[U] 20m/s.
What will be the maximum height[H] attained by the ball?
V2=U2+2gS
v=o,U=20m/s,g=-10m/s2
0=400+2*[-10]*S
400=-20S
S=400/-20=20m
S=H=20m
Then what will be the time [t] taken by the ball to reach 20m or H ?
V=U+gt
0=20+-10t
20=-10t
t=20/10=2 sec
If we tosses up a ball with an initial velocity 20m/s it will attain a maximum height 20 m in 2 seconds.

After 2 second its final velocity[V] becomes zero.Then ball starts a free fall.
What will be the distance travelled by a free falling body after one second?
S=Ut+1/2gt2,0+1/2*10*1=5m
So the ball travelled 5m downwards,say ,that point as 'A'.From 'A' ball has to travel 15m more to reach the ground.

Now consider the second ball,it tosses up ,when the first ball reaches 20m[H].We know that second ball also attain 20 m at 2 second.

The first ball[after 1 second] is now at 'A' [5m or1/4H],it is travelling down wards .
Then what may be the position of the second ball after 1 second?[Going up wards]

S=Ut+1/2gt2
S=20*1+1/2[-10]*1
=20+-5=15m or 3/4H or At point 'A'
Now After one second, first ball [downwards 5m]and second ball [up wards 15m or 3/4H]may collide at point 'A'
Point 'A' is at 15 th meter if H is 20 meter .ie,the answer is 3/4H.

Let us discuss some more interesting things,related to this, like escape velocity , After your comments .

thank u 4 reading me patiently.