Apoliceman 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)

They will meet at a distance of 7.57 m

Given:

Initial velocity of policeman in the x- direction,

The distance between the buildings,

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:

t = 0.707 s

Now,

When the policeman was chasing across:

The distance they will meet at:

9.57 - 2.0 = 7.57 m

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.

b. false

explanation:

suppose a rock in free fall is dropped off a cliff. after falling for 5 seconds, its speed would not be 100 m/s.