Solving for Time: Dropping a Brick from a Building

[saiyajin822]

more kinetics :(

sigh here's another one i thought i knew how to do but i ended up with the wrong answer.

A brick is dropped from the roof of a building. Twenty meters below the point of release is the top of a window. How long will it take the brick to pass in front of the window which is 21.5 m tall?

d=21.5 m
a=9.8m/s
vi=0
t=?
i used d=vit+1/2a(t)^2 formula so..

21.5=0+1/2(9.8)(t)^2

did i do that right...

 

[Dark_Adonis]

sigh here's another one i thought i knew how to do but i ended up with the wrong answer.

A brick is dropped from the roof of a building. Twenty meters below the point of release is the top of a window. How long will it take the brick to pass in front of the window which is 21.5 m tall?

d=21.5 m
a=9.8m/s
vi=0
t=?
i used d=vit+1/2a(t)^2 formula so..

21.5=0+1/2(9.8)(t)^2

did i do that right...

I'm afraid you may not have. I suggest that instead you might take the v_i as the velocity of the brick at the point where the brick passes the top of the window. (d=20m) Then your equation would work, but I'm just an idiot...
I thought about this for a second and realized a simpler method would be to simply take the difference in times of d=20m and d=41.5m. (Use your original formula set up and solve for time when d=20m then solve when d=41.5, then find the difference)

 

[recon]

I don't like using formulae, and since I've not been taught more complicated kinematics at school yet, I'll do it the non-formulae way.

The brick has initial velocity of 0 m/s. When dropped, it accelerates towards the ground at 9.8 m/s^2. We want to find its velocity at the point where it reaches the top of the window. Visualise a velocity-time graph in your head. From the point of origin, you can draw a line that has gradient 9.8. We want to calculate just how far we can draw the line before the area underneath the line is 20m. Let's call the time needed to travel this distance t₁. Then the velocity at this point will be 9.8t₁. The area underneath the line is 1/2 x 9.8t₁ x t₁ = (9.8t₁ ^2)/2.

So,
9.8t₁ ^2 = 40
t₁ = 2.020305089

Use the same method to determine the time taken for the brick to pass through both the 20 m before reaching the window and the height of the window itself (21.5m) which has a total distance of 41.5m. Call this time t₂. Minus t₁ from t₂, and you'll have your answer.

 

[JasonRox]

I hate graphs. They are very annoying.

I'll plot them, but I just hate those who ask plot the following when x equals:

.9
.99
.999
.9999
1.1
1.01
1.001


It never ends!

 

[saiyajin822]

WOW i mistyped the question, I am really sorry about that. the window is actually 1.5 m high. This is what the real problem should say:

A brick is dropped from the roof of a building. Twenty meters below the point of release is the top of a window. How long will it take the brick to pass in front of the window which is 1.5 m tall?

again,sorry

by the way, i did it the way u said but i don't understand why we would take the difference of using 20m and 21.5m.

 

[recon]

I hate graphs. They are very annoying.

I'll plot them, but I just hate those who ask plot the following when x equals:

.9
.99
.999
.9999
1.1
1.01
1.001


It never ends!

I did not say 'plot a graph'. I said visualise the graph in your head, and you'll be able to solve the problem.

 

[recon]

WOW i mistyped the question, I am really sorry about that. the window is actually 1.5 m high. This is what the real problem should say:

A brick is dropped from the roof of a building. Twenty meters below the point of release is the top of a window. How long will it take the brick to pass in front of the window which is 1.5 m tall?

again,sorry

by the way, i did it the way u said but i don't understand why we would take the difference of using 20m and 21.5m.

Even though the value has now changed, the method used in solving the question remains the same. Hence, if you understood my method for solving the initial problem, you should be able to tweak it to work for this problem as well.

In your original question, t₁ is the time taken to fall from the roof to the top of the window. t₂ is the time taken to fall from the roof to the bottom of the window. Hence to find the time taken to fall from the top of the window to the bottom of the window, you need to find the difference between t₁ and t₂.