Solving a Problem: Calculating Brick Velocity, Height and Time

[Alain12345]

I solved a problem my teacher assigned, but I don't know if I did it right. The question is:

A person standing on the roof of a building throws a brick straight down at 7 m/s. THe brick hits the ground 4 seconds later.
Calculate:
a) The velocity at which the brick hits the ground (I got 46.24 m/s)
b) The height of the building (I got 106.48 m)
c) If the brick is thrown upward instead, at 7 m/s, how much longer will it take to hit the ground? (I said it would take 11.108 s)

Thanks.

 

[physhelp]

what did you receive for time? 7m/s is initial velocity right?

 

[Alain12345]

It hit the ground 4 seconds after, so that's the time... and 7 m/s is the initial velocity.

 

[dicerandom]

I agree with (a) and (b), but I got a different number for (c).

 

[Alain12345]

What was your number? I got mine by getting the time it traveled up at 7m/s. It hit zero velocity, and started accelerating towards the ground by the acceleration due to gravity (9.81 m/s2). Then I calculated how many seconds it took from zero velocity to the ground. Is my method even correct?

Edit: Sorry, I wrote down what I did wrong... this isn't the method I used...

 

[dicerandom]

Well I don't want to give you the answer

You need to use the formula:

x(t) = \frac{1}{2} a t^2 + v_0 t + x_0

and pick appropriate values for a, v_0, and x_0. Your specific values will depend on your coordinate system, of course. Just make sure that you're explicit with where the origin of your coordinate system is and which way you're calling positive, then make sure that all your numbers are consistent with that coordinate system.

 

[dicerandom]

for the height of the building I got 52.9, i might be wrong but this is how i got mine. X=0, A=-9.8, t=4, Vo=7...i solved for Xo in the formula X=1/2at^2+volt+Xo

If you're saying that the acceleration is negative then you're saying that gravity pulls the brick in the negative x direction. This means that your initial velocity should also be negative.

Like I said above, it's important to be very careful about what your coordinate system is and make sure that all the values you're using make sense in that coordinate system. That's really the hardest part with these problems is making sure that you're consistent.