please answer it quick it's urgent From the top of a tower 100 m in height a ball is dropped and at the same time
another ball is projected vertically upwards from the ground with a velocity of 25 m/s.
Find when and where the two balls will meet. (g = 9.8 m/s)
Answers
The two balls will meet at 80.4 m from ground after 4 seconds.
Solution:
Let us take distance covered by the stone thrown upwards to meet the other as x. Then the other stone covers a distance of (100 - x) m
Let the distance covered by the ball dropped down be b1 and the distance covered by the ball thrown up be b2.
b1 = d = 100-x
g\quad =\quad 9.8\quad m/s^{ 2 }g=9.8m/s
2
u = 0
We know, s\quad =\quad ut\quad +\quad \frac { 1 }{ 2 } at^{ 2 }s=ut+
2
1
at
2
Putting values,
100\quad -\quad x\quad =\quad 4.9t^{ 2 }\quad \longrightarrow (1)100−x=4.9t
2
⟶(1)
b2\quad \Rightarrow \quad d\quad =\quad xb2⇒d=x
g\quad =\quad 9.8\quad \frac { m }{ s^{ 2 } }g=9.8
s
2
m
u\quad =\quad 25\quad \frac { m }{ s }u=25
s
m
We know, s\quad =\quad ut\quad +\quad \frac { 1 }{ 2 } at^{ 2 }s=ut+
2
1
at
2
Putting values,
x\quad =\quad 25t\quad -\quad 4.9t^{ 2 }\quad \longrightarrow (2)x=25t−4.9t
2
⟶(2)
Adding (1) and (2),
100\quad -\quad x\quad +\quad x\quad =\quad 4.9t^{ 2 }\quad +\quad 25t\quad -\quad 4.9t^{ 2 }100−x+x=4.9t
2
+25t−4.9t
2
t = 4 s
Putting t = 4 in (1),
100\quad -\quad x\quad =\quad 4.9t^{ 2 }100−x=4.9t
2
100\quad -\quad x\quad =\quad 78.4100−x=78.4
x = 80.4 m"