Question

A policeman is chasing a criminal across a rooftop at 10 m/s. He decides to jump to the next building which is 2 meters across from the original and 2.5 meters lower. How far (in m) will they fall in that time? (use g = 10 m/s^2)

Answer 1

Answer:

They will meet at a distance of 7.57 m

Given:

Initial velocity of policeman in the x- direction, $u_{x} = 10 m/s$

The distance between the buildings, $d_{x} = 2.0 m$

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:

$h = ut + \frac{1}{2}gt^{2}$

$h = \frac{1}{2}gt^{2}$

$2.5 = \frac{1}{2}\times 10\times t^{2}$

t = 0.707 s

Now,

When the policeman was chasing across:

$d_{x} = u_{x}t + \frac{1}{2}gt^{2}$

$d_{x} = 10\times 0.707 + \frac{1}{2}\times 10\times 0.5 = 9.57 m$

The distance they will meet at:

9.57 - 2.0 = 7.57 m