Answer:
6.17 m
Explanation:
We are given that
Initial speed of rubber ball, u=12 m/s
Total height, h=7 m
Initial speed of second ball, u'=0
We have to find the height at which the two balls collide.
Let first rubber ball and second ball strikes after t time.
For first ball
Distance traveled by first ball in time t
[tex]S=ut-\frac{1}{2}gt^2[/tex]
Substitute the value
[tex]S=12t-\frac{1}{2}(9.8)t^2[/tex]
[tex]S=12t-4.9t^2[/tex] ...(1)
Distance traveled by second ball in time t
[tex]7-S=\frac{1}{2}(9.8)t^2[/tex]
[tex]7-S=4.9t^2[/tex] .....(2)
Using equation (2) in equation (1) we get
[tex]S=12t-(7-S)[/tex]
[tex]S=12t-7+S[/tex]
[tex]\implies 12t=7[/tex]
[tex]t=\frac{12}{7}[/tex]sec
Now, using the value of t
[tex]S=12(\frac{12}{7})-4.9(\frac{12}{7})^2[/tex]
S=6.17 m
Hence, at height 6.17 m the two balls collide .
