
#1
Sep3007, 10:49 PM

P: 39

1. The problem statement, all variables and given/known data
An object is thrown up from the top of a building 50 m high. It rises to a maximum high of 20 m above the roof. A) when does it land on the ground B) At what velocity does it land ? C) When is it 20 m below the roof? 2. Relevant equations The 4 kinematics equations the ones I used where Vf = Vi+ at vf ² = vi ² + 2ax 3. The attempt at a solution Well I am completely lost now. I don't know if we can do this in one shot or two. I first found out how much time it takes for the ball to attein maximum height which gave me 2.02 seconds. After I calculate the time it takes to go from that high to the ground and in total I got 6.35 s but the textbook gave me 5.84 as for the velocity I had  42 m/s but the text book says 37 m/s :S I kept alot of decimals so I don't think the mistake is there 



#2
Sep3007, 11:18 PM

P: 1,877

So for the first path you should get that
[tex] t_1 = \frac{\sqrt{2gy_{high}}}{g}[/tex] For the second path you should use the distance equation [tex] t_2 = \sqrt{\frac{2 y_{net}}{g}}[/tex] These two times will give you just about everything. Is this what you got? 



#3
Sep3007, 11:26 PM

P: 48

Step 1 : Decide the reference point for measuring the displacements i.e. y = 0 point and the +ve direction for displacement. It could be the ground in which case the initial displacement will be +50m and final displacement 0. Alternatively it could be the top of the building with upwards displacement +ve. In that case initial displacement will be 0 and final displacement 50m. Let us assume the second alternative.
Remember : velocities and accelerations in the direction of +ve displacement are +ve and otherwise ve. Step 2 : First determine u by using v^2  u^2 = 2as. At the highest point v = 0. Displacement s = +20m and acceleration a = g = 9.8 m/s2 (ve because g is downwards). Step 3 : Use s = ut + 1/2 at^2 to determine time of hitting the ground. s = 50m and a = g Step 4 : To determine the landing velocity use v^2  u^2 = 2as. u as determined above, a = g and s = 50m Step 5 : When is it 20 m below the roof : s = 20m. Which equation will you use? 



#4
Sep3007, 11:29 PM

P: 1,877

Kinematic problem
Good advice by Vijay, the only thing that I would add is to not put in any numbers until the absolute last moment possible. Do everything algebraically.




#5
Sep3007, 11:31 PM

P: 39

Hmmm am not so familiar with those symbols is it,
u = initial velocity Still looking at it for now 



#6
Sep3007, 11:50 PM

P: 39

Well thank you very much , this was pretty helpful I was never using a reference point correctly before, just got the last answer with Vf=Vi +at



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