Answer:
They will meet at a distance of 7.57 m
Given:
Initial velocity of policeman in the x- direction, [tex]u_{x} = 10 m/s[/tex]
The distance between the buildings, [tex]d_{x} = 2.0 m[/tex]
The building is lower by a height, h = 2.5 m
Solution:
Now,
When the policeman jumps from a height of 2.5 m, then his initial velocity, u was 0.
Thus
From the second eqn of motion, we can write:
[tex]h = ut + \frac{1}{2}gt^{2}[/tex]
[tex]h = \frac{1}{2}gt^{2}[/tex]
[tex]2.5 = \frac{1}{2}\times 10\times t^{2}[/tex]
t = 0.707 s
Now,
When the policeman was chasing across:
[tex]d_{x} = u_{x}t + \frac{1}{2}gt^{2}[/tex]
[tex]d_{x} = 10\times 0.707 + \frac{1}{2}\times 10\times 0.5 = 9.57 m[/tex]
The distance they will meet at:
9.57 - 2.0 = 7.57 m
