EK phys 1001 proportion problem

[Jonass00]

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I am having some difficulty with proportions when they involve square roots and exponents.

Problem #102. Object dropped from H lands with velocity V. If object dropped from 2H what represents the velocity when it strikes the ground.

I used V= Sqrt 2GH. I understand that by increasing the H by 2, it has an overall effect of increasing the V by 1.4 whichSqrt 2. simply inserting a 2 under the sqr root would increase the V by sqr root of 2.


In the following question #103.

Object is dropped from height H and strikes ground in time T. If object is dropped from object 2H which is the following represents time it wil take to strike ground?

using the shortened H= 1/2at^2. Answer is 1.4 once again. My thought process gets mixed up a little here. by increasing 2H, does that also mean multiplying T by 2? which would end up giving you 4t (my answer choice).

However in #104... Object dropped from height H and strikes ground in time T. In order to double flight time, object must be dropped from height of? answer is 4H. My thought process resulted with answer of 2H. 2H would mean multiplying 2 to T on the right side. obviously this is incorrect. Anyone care to help and see where I am failing to make the connection. Thank you in advance.

obviously I am not making the connection here as to how they are answering this proportion question. Anyone care to chime in.

 

[akim6890]

From your equation of H = 1/2at^2, rework it to t = sqrt(2H/a) and the proportion should be obvious from there. To increase time by 2, increase H by 4. Remember that H is proportional to t^2 not 2t. So if they asked you to triple the time of flight, the height would be 9H.

 

[Gauss44]

I am having some difficulty with proportions when they involve square roots and exponents.

Problem #102. Object dropped from H lands with velocity V. If object dropped from 2H what represents the velocity when it strikes the ground.

I used V= Sqrt 2GH. I understand that by increasing the H by 2, it has an overall effect of increasing the V by 1.4 whichSqrt 2. simply inserting a 2 under the sqr root would increase the V by sqr root of 2.


In the following question #103.

Object is dropped from height H and strikes ground in time T. If object is dropped from object 2H which is the following represents time it wil take to strike ground?

using the shortened H= 1/2at^2. Answer is 1.4 once again. My thought process gets mixed up a little here. by increasing 2H, does that also mean multiplying T by 2? which would end up giving you 4t (my answer choice).

However in #104... Object dropped from height H and strikes ground in time T. In order to double flight time, object must be dropped from height of? answer is 4H. My thought process resulted with answer of 2H. 2H would mean multiplying 2 to T on the right side. obviously this is incorrect. Anyone care to help and see where I am failing to make the connection. Thank you in advance.

obviously I am not making the connection here as to how they are answering this proportion question. Anyone care to chime in.

#104
X=.5at^2 so
X is proportional to T^2
T=sqrtX

Plug in numbers: sqrt of 4 times distance = 2 times time

 

[noobtech]

#103> 1/2 a t^2 = H --> t = sqrt(2h/a) ... so when you have 2H, it's multiplied by sqrt(2)
#104> t = 2sqrt(2h/a) so, new height is multiplied by 4