Projectile Motion: Solving for Horizontal Displacement from a 200m Cliff

[jlm342000]

Homework Statement

A projectile is fired into the air from the top of a 200 m cliff above a valley as shown in the figure. Its initial velocity is 60 m/s above the horizontal. Where does the projectile land?

Homework Equations

x=60cos60 t

set the next equation equal to zero and solve for t:

-200=60sin60t-1/2(-9.81)(t)^2

The Attempt at a Solution

I solved for t so I could find horizontal displacement, ended up with the following quadratic but end up with a negative number inside square root.
4.9t^2+52t+200=0 I know my mistake is probably obvious but I can't see it, help appreciated.

 

[rl.bhat]

-200=60sin60t-1/2(-9.81)(t)^2
In the equation you have to use only one negative.

 

[ideasrule]

The equation is displacement=v0t+(1/2)at^2. You have to choose one direction as positive and write all quantities according to that convention. If you do that, you'll end up with only one negative, as rl said.

 

[jlm342000]

-200=60sin60t-1/2(-9.81)(t)^2
In the equation you have to use only one negative.

But isn't gravity negative? So don't I plug a negative value here?

 

[jlm342000]

Appreciate it

 

[ideasrule]

No, you always use the equation:

displacement=v0t+(1/2)at^2

If you choose "up" as positive, then displacement would be -200, which you have. v0 would be vsin(60), which you also have. a would be -9.8 m/s^2, since gravity acts downwards. Leave the positive sign in front of 1/2 alone; that can never change.