Projectile Motion, how tall is the building

[xRantastic]

Homework Statement

A ball thrown horizontally at 21.7 m/s from the roof of a building lands 35.5 m from the base of the building. How tall is the building?

Homework Equations

The Attempt at a Solution

Vy= Voy + 1/2AT^2
Vy= -4.9(21.7)^2
Vy= -2307.361

I know it should be positive but MasteringPhysics.com keeps telling me that it is wrong whether positive or negative!

:|

 

[thegreenlaser]

I'm not really sure what you plugged into that formula... From what I can see you plugged the horizontal velocity in for 'time' and then solved for the vertical velocity. I don't think that will work even as far as units go.

I don't think you can solve directly for the vertical displacement with what you're given, however, you're given horizontal displacement, and initial horizontal velocity. Now, what's the horizontal acceleration? (shouldn't need any calculation) Knowing that, how would you solve for the time that the ball spends in the air? Then, knowing the change in time, and knowing the vertical acceleration (due to gravity) you should be able to solve for the vertical displacement.

 

[xRantastic]

Okay..heres what i did

X=VoxT --> T=Vox/x

T= 35.5/21.7
T=1.64

Thats its horizontal acceleration.

But are you sure that by multiplying that by -9.8 will give me the vertical displacement?

 

[xRantastic]

Thanks a lot greenlaser! Really appreciate it.

 

[gneill]

Okay..heres what i did

X=VoxT --> T=Vox/x

T= 35.5/21.7
T=1.64

Thats its horizontal acceleration.

But are you sure that by multiplying that by -9.8 will give me the vertical displacement?

No, T in the above is time; the time it takes for the projectile to reach distance 35.5m given that its velocity in the x-direction is 21.7 m/s.

While the projectile is traveling horizontally for time T, it is also falling vertically (separate component of the overall motion!). The question is, how far will it fall vertically in the same time T?