Question

A cement block accidentally falls from rest from the ledge of a 53.4-m-high building. When the block is 19.4 m above the ground, a man, 2.00 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way

Answer 1

The time required by the man to get out of the way is 0.6 s.

Step-by-step explanation:

height of building, H = 53.4 m

height of block, h = 19.4 m

height of man, h' = 2 m

Let the velocity of the block at 19.4 m is v.

use third equation of motion

`v^2 = u^2 + 2 gh\\\\v^2 = 0 + 2 * 9.8 * (53.4 - 19.4)\\\\v = 25.8 m/s`

Now let the time is t.

Use second equation of motion

`h = u t + 0.5 gt^2\\\\19.4 - 2 = 25.8 t + 4.9 t^2\\\\4.9 t^2 + 25.8 t - 17.4= 0 \\\\t = (-25.8\pm√(665.64 + 341.04))/(9.8)\\\\t = (-25.8\pm31.7)/(9.8)\\\\t = 0.6 s, - 5.9 s`

Time cannot be negative so time t = 0.6 s.