Respuesta :
Answer:
The man has, at most, 0.4 secs to get out of the way
Explanation:
First, we will determine, the time the block takes to reach height 13.4 m above the ground.
Also, we will determine the time it will take to reach the man, if he does not notice it.
Then, the difference in the two times will be the time the man has to get out of the way.
From the formula
[tex]h =ut - \frac{1}{2}gt^{2}[/tex]
Where [tex]h[/tex] is height
[tex]u[/tex] is the initial velocity
[tex]t[/tex] is time
and [tex]g[/tex] is acceleration due to gravity ( Take [tex]g =[/tex] 9.8 m/s²)
Let the time the block takes to reach height 13.4 m be [tex]t_{1}[/tex]
The cement block falls from a 51.1 m high building, then by the time it reaches a height 13.4 m above the ground, the displacement is
13.4 m - 51.1 m = - 37.7 m
The is the height the block has fallen from by the time it gets to a height 13.4 m above the ground
Then,
[tex]h =ut - \frac{1}{2}gt^{2}[/tex]
[tex]-37.7 = (0)(t_{1}) - \frac{1}{2}(9.8)t_{1}^{2}[/tex]
(NOTE: [tex]u[/tex] = 0 m/s because the cement block falls from rest)
[tex]37.7 = 4.9t_{1}^{2}[/tex]
[tex]t_{1}^{2}= \frac{37.7}{4.9}[/tex]
[tex]t_{1}= \sqrt{7.69} \\[/tex]
[tex]t_{1}= 2.77 secs[/tex]
Hence, It takes the block 2.77 secs to reach height 13.4 m
Also,
Let the time it will take to reach the man be [tex]t_{2}[/tex]
The man is 1.70 m tall, hence, to reach the man, the cement block would have reached a height ( 1.70 m - 51.1 m) = - 49.4 m
Then,
[tex]h =ut - \frac{1}{2}gt^{2}[/tex]
[tex]-49.4 = (0)(t_{2}) - \frac{1}{2}(9.8)t_{2}^{2}[/tex]
[tex]49.4 = 4.9t_{2}^{2}[/tex]
[tex]t_{2}^{2} = \frac{49.4}{4.9}[/tex]
[tex]t_{2}= \sqrt{10.08}[/tex]
[tex]t_{2}= 3.17 secs[/tex]
Hence, it will take 3.17 secs to reach the man
Now, for the time the man has to get out of the way, that is
[tex]t_{2} - t_{1}[/tex] = 3.17 secs - 2.77 secs
= 0.4 secs
Hence, the man has, at most, 0.4 secs to get out of the way
