Question

A cement block accidentally falls from rest from the ledge of a 52.9-m-high building. When the block is 14.3 m above the ground, a man, 1.94 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

Answer:

The man has at most 0.418 secs to get out of the way

Explanation:

To determine how much time at most the man has to get out of the way, we will calculate the time it will take the block to reach height 1.94m from height 14.3m.

To do this, we will first determine the time it will take the block to reach height 1.94 m from height 52.9 m and find the time it takes the block to reach height 14.3m above the ground from the same height (52.9 m), the difference is the time the man has to get out of the way.

Now, the time it will take the block to reach height 1.94 m from height 52.9 m

This means the time it will take the block to travel a height distance of 52.9m - 1.94m = 50.96m

From one of the equations of motions for free falling bodies

h = ut + 1/2(gt²)

Where h is the height

u is the initial velocity

t is the time

and g is the acceleration due to gravity (Take g = 9.8 m/s²)

From the question, the block falls from rest

∴ u = 0 m/s

h = 50.96 m

Putting these into the equation

50.96 = 0(t) + 1/2(9.8)(t²)

50.96 = 4.9t²

t² = 50.96/4.9

t² = 10.4

t = √10.4

t = 3.225 secs

This is the time it will take to reach height 1.94m (that is to reach the man)

For the time it takes the block to reach height 14.3m above the ground from height 52.9 m

That is, the time it takes the block to travel a height distance of 52.9m - 14.3m = 38.6 m

Here,

h = 38.6 m

and u = 0 m/s

Putting these into the same equation

h = ut + 1/2(gt²)

38.6 = 0(t) + 1/2(9.8)(t²)

38.6 = 4.9t²

t² = 38.6/4.9

t² = 7.878

t = √7.878

t = 2.807 secs

This is the time it takes the block to reach height 14.3 m

Now, the difference in time is 3.225secs - 2.807 secs = 0.418 secs

Hence, the man has at most 0.418 secs to get out of the way.