13. A ball which is thrown vertically upwards reaches the roof of a tower 100 m high. At the moment
this ball is thrown vertically upward, another ball is dropped from rest vertically downwards from
the roof of the tower. Calculate when and where these balls will meet each other? (g = 9.8 m/s)
Ans. After
100/14√2
second. At a height of 75 m from the earth
Answers
yes, your answer is correct he balls pass each other at a height of 75m
just in case you want an intensely detailed answer so
First find the initial velocity the first ball had when it was thrown vertically up:
at max height v(f) = 0 m/s
a = -g = -9.8 m/s² ... [taking UP as the positive direction]
s = 100 m
v(f)² = v(i)² + 2as
v(i)² = v(f)² - 2as
v(i) = √[0 - 2 * -9.8 * 100]
v(i) = 44.27 m/s
When the two balls meet their distances above the level the first ball was thrown from (assumed to be the ground ... b/c no other info has been given about it) are equal
The 2nd ball is dropped at the same time the 1st ball is thrown ... so the times taken to the meeting point for the balls are also equal
For the 2nd ball the initial vertical velocity is 0 m/s ... [b/c it is just dropped]
s = v(i)t + (1/2)at²
so for the 1st ball:
s = 44.27t - 4.9t²
and for the second ball:
-s = 0 - 4.9t² ... [-s b/c the 2nd ball travels in the negative direction]
so s = 4.9t²
BUT we want the distance the 2nd ball is above the ground so
height the 2nd ball is above the ground when the two balls meet = 100 - 4.9t² ... [eqn 1]
then when the two balls meet:
44.27t - 4.9t² = 100 - 4.9t²
44.27t = 100
t = 100/44.27
subs t = 100/44.27 into [eqn 1]:
distance above the ground where the two balls meet = 100 - 4.9 * (100/44.27)² = 75 m
so the balls pass each other at a height of 75m
Let ball 1 be the ball thrown vertically upward and ball 2 be the ball thrown vertically downward.
Height of tower is = 100 m
=> First , we will find out initial velocity of ball 1.
We know that :- v²-u²=2as
where :-
- v = final velocity
- u = initial velocity
- a= acceleration
- s= distance
Final velocity of ball 1 will be zero metre per second because at highest point velocity is zero.
Let initial velocity of ball1 be m
=> v²-u²=2as
=> 0²-m²=2×(-g)×100
=> -m²=-2×9.8×100
=> -m²=-2×(98/10)×100
=> -m²=-2×98×10
=> m²=2×98×10
=> m²=2×98×10
=> m²= 1960
=> m= √(1960 )
=> m= 14√10 m/s
Now let's assume that, both the balls will meet at point X. Ball 1 is at point A and ball 2 is at point B.
Let distance between point B and X be h metres.
Therefore, distance between point X and A =( 100-h ) metres
We know that :- s= ut+(1/2 )at²
For ball2 , initial velocity is zero.
Let time taken by ball 2 to reach at point X be T
=> s= ut+(1/2 )at²
=> h= (0×t)+(1/2 )gT²
=> h= (1/2 )×9.8× T²
Now we will put values for ball 1 in the equation s= ut+(1/2 )at². Since the balls will meet at point X , so time taken by ball 1 to cover distance 100- h will also be T.
=> s= ut+(1/2 )at²
=> 100-h = 14√(10)T-(1/2 )gT²
=> 100-h = 14√(10) T-[(1/2 )× 9.8×T²]
But , h= (1/2 )×9.8× T²
=> 100-h = 14√10 T- h
=> 100-h+h = 14√10 T
=> 100 = 14√10 T
=> 100 / (14√10) seconds= T
or
=> 100 / √(1960) seconds= T
Now we will find h.
=> h= (1/2 )×9.8× T²
=> h= (1/2 )×9.8× [100 / √(1960) ]²
=> h= (1/2 )×(98/10)×( 100×100 / 1960 )
=> h= (1/2 )×98×( 100 / 196 )
=> h= (1/2 )×1×( 100 / 2 )
=> h= 100 / 4
=> h= 25 m
Now , from the ground level ball will meet at (100-h) metres.
=> 100-h
put h = 25
=> 100- 25 = 75 metres
[ Kindly refer attachment for diagram ]
