Solving Kinematic Problem: Object Thrown From 50m High Building

[EvilBunny]

Homework Statement

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?

Homework Equations

The 4 kinematics equations the ones I used where

Vf = Vi+ at

vf ² = vi ² + 2ax

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 textbook says -37 m/s :S


I kept a lot of decimals so I don't think the mistake is there

 

[Mindscrape]

So for the first path you should get that

$$t_1 = \frac{\sqrt{2gy_{high}}}{g}$$

For the second path you should use the distance equation

$$t_2 = \sqrt{\frac{2 y_{net}}{g}}$$

These two times will give you just about everything. Is this what you got?

 

[Vijay Bhatnagar]

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?

 

[Mindscrape]

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.

 

[EvilBunny]

Hmmm am not so familiar with those symbols is it,

u = initial velocity

Still looking at it for now

 

[EvilBunny]

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