Finding speed of a falling brick using distance and the acceleration of gravity

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The discussion centers on calculating the speed of a falling brick after it has dropped 30 meters under the acceleration of gravity at 10 m/s². The initial calculation of speed as 10 m/s is incorrect, as it confuses average speed with instantaneous speed. Participants emphasize the importance of the SUVAT equations for solving such physics problems. The original poster expresses confusion but later acknowledges understanding after receiving guidance on these equations. The conversation concludes with a positive note of gratitude for the assistance provided.
vytrx
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Homework Statement
A supervisor 1.8m tall visits a construction site a brick resting at the edge of a roof 50m above the ground suddenly falls at the instant when the brick has fallen 30m without encountering any air resistance, the supervisor sees the brick coming down directly towards him from above as shown in the following figure. (attached) Calculate the speed of the brick after it has falled 30 m from the roof
Relevant Equations
Speed=distance/time
acceleration of gravity is 10m/s^2
Since the brick has fallen 30m and the acceleration of gravity is 10m/s^2 the brick would have fallen 3 seconds speed of brick would then be 30/3 leaving the answer at 10m/s?I am new to physics and this question has left me stumped
 

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vytrx said:
Relevant Equations:: Speed=distance/time
That's only true for constant speed. Or, in simple cases, for calculating the average speed.

Do you know the SUVAT equations?

vytrx said:
Since the brick has fallen 30m and the acceleration of gravity is 10m/s^2 the brick would have fallen 3 seconds
You may be confusing speed and acceleration here.
vytrx said:
speed of brick would then be 30/3 leaving the answer at 10m/s?
This is not correct.

:welcome:
 
PeroK said:
That's only true for constant speed. Or, in simple cases, for calculating the average speed.

Do you know the SUVAT equations?You may be confusing speed and acceleration here.

This is not correct.

:welcome:
No I'm not familiar with the SUVAT equations, yes I may be confusing the two would you care to explain how this question is done? I've asked one of my friends that said that the answer was 24.5 but I'm not entirely sure how to get there
 
vytrx said:
No I'm not familiar with the SUVAT equations, yes I may be confusing the two would you care to explain how this question is done? I've asked one of my friends that said that the answer was 24.5 but I'm not entirely sure how to get there
The SUVAT equations are the starting point for a study of physics, so you'll need to learn about them. Try this to get you started.

https://www.ncl.ac.uk/webtemplate/a...mechanics/kinematics/equations-of-motion.html
 
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Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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