Height of a building with a Projectile

[guma671]

Homework Statement

A ball is launched at an angle of 38° at 16m/s off of the top of a building, the ball finally hits the ground after 45seconds. What is the height of the building? Gravity is normal 9.80m.s squared. The answer to the problem is 9480m.

Homework Equations

I just needed a starting point. I wasn't sure really where to go with the problem.

The Attempt at a Solution

First i tried to find Vy by 16.0m/s * sin(38°) and got a value of 4.741m/s. I then plugged it into the equation Δy=VyiΔt+1/2gΔtsquared but the answer came out to around 10,000.

 

[SteamKing]

Why don't you show your complete calculations?

 

[guma671]

what do you mean?

 

[guma671]

oh well with the answer i got around 10,000 it was just Y=(16m/s)(45s)+1/2(-9.8)(45)squared which came out to -9202.5

 

[BvU]

Always good to check things.
1. You know sin(30°) = 0.5 so you should expect Vy > 8 m/s !
2. If even that leads to an improbable answer, check that you read the OP correctly. Perhaps it says 4.5 seconds ?

 

[BvU]

Posts crossed. I am a slow typer. Now you use Vo, not Vy. Not that it matters much on a 10000 m scale. But it does on a 100 m scale.

 

[BvU]

Come again. What building is 9480 m high ?

 

[BvU]

If the 9480 m is in the answer book, you may have to deal with a case of sloppy re-use of old exercises. They change a number but don't bother to check if it's sensible. In this case that is double painful: the 9480 is nonsense and on top of that, after 1 second, the vertical component of the velocity is 0. In the 44 seconds after that it would go -430 m/s, way over the speed of sound. If the building is on Earth (9.8 m/s^2), the air restance can't be ignored anymore, even for a lead ball.