ഫിസിക്സ് അദ്ധ്യാപകന്‍: December 2011

QUESTION BOX, PYARILAL SIR EXPLAINS........

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

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

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

thank u 4 reading me patiently.

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