Ball launched off a cliff at negative angle

[picklepie159]

Ball launched off a cliff at negative angle...

Homework Statement

So a ball is launched off a cliff 60 meters high
It is thrown at -15 degrees horizontal, and lands 120 meters away.

Find the V-initial, V-final, Time in air, and Angle at which it hits the ground

Homework Equations

inverse tan(75)= Vx/Vy
D= VT
h=1/2g t^2
Vx= cosin(75) x V-initial
Vy=sin(75) x V-initial

The Attempt at a Solution

I know that there's a policy that I have to try to work on the problem first before i get help, but man, I am lost. This is what I can make out.

I switched everything to positive so now its falling off the cliff at 75 degrees.
120 meters = Vx T

That's all. A little ray of guidance would be a Godsend.

PS
Would the Velocity-y be V-initial +34.6 m/s?
Because when something falls 60 meters from rest, the final velocity is 34.6 meters.

 

[Pengwuino]

What do you mean you switched everything to positive?

 

[The legend]

switching everything positive is something i don't understand in your solution...

so my approach would be ...
Range = v*cos(-15) * t

Height = v*sin(-15)t + 1/2*g*t^2

now ... you get 2 equations in v and t and there are 2 variables... so you can solve for both...do you get it?

 

[picklepie159]

By switching to positive, I ust meant I flipped it upside down- the movement towards the ground is now the positive direction. Then, the angle it was thrown at is 75 degrees instead of -15. (Does that work out?)

It just makes stuff easier- no fiddly negative directions to deal with.

 

[The legend]

then don't you think the time of flight changes in both cases?

 

[Nightro]

Really sorry for digging up a super old thread but I'm trying to solve a similar problem right now.

As The Legend suggested, I've solved for v and t using those equations. First, I define v in terms of R and t using the R = v * cosθ * t

v = R/(cos(θ) * t)

Then, I substitute that into the height equation...

H = (R/(cos(θ) * t)) * sin(θ) * t + 1/2 * g * t^2

and solve for t...

t = √((R * tan(θ) + H)/(g * 1/2))

Solve it using my values, then substitute that back into the velocity derivation above...

This works for angles [0..90] but anything below the horizon just craps out due to a -ve √

Where am I going wrong? This is driving me insane!

 

[Nightro]

Just wanted to clarify something, the reason I add H is because I actually got

--H in my particular case..

t = √((R * tan(θ) + H)/(g * 1/2))

 

[Nightro]

Actually... I'm wrong.. it works for the inputs given at the top of this thread :S

I think what might actually be happening in my case is my launch angle is too much!

That is to say... the angle that is a direct line to the target location is more than the angle I'm trying to launch it at (e.g.)

Angle from Launch to Target: -10
Launch Angle: -20

That would mean its impossible, right? And why I'm getting these imaginary numbers :)