anonymous
  • anonymous
I am a little confused on how to solve this, could someone help me out? A cement block accidentally falls from rest from the ledge of a 73.2-m-high building. When the block is 17.0 m above the ground, a man, 1.70 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?
Physics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Here initial vel, u=0 as it falls from rest accl, a=-g=-9.8 So \[S=u*t+0.5*a*t^2\] becomes \[S=-0.5*9.8*t^2\]
anonymous
  • anonymous
Time to reach 17m, t1: Here displacement, S=17-73.2 t=t1 Substitute in the above eqn and find t1...
anonymous
  • anonymous
Time to reach 1.7m, t2: S=1.7-73.2 ; t=t2 find t2....

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Time of travel from 17m to 1.7 m= t2-t1 He must evade before this time to save himself...

Looking for something else?

Not the answer you are looking for? Search for more explanations.