Calculating initial trajectory without time

[Latsabb]

Ok, I am a bit stuck on a homework problem. I know the following:

Initial speed: 25 m/s
Distance: 49.85 m
Initial height: 30 m

And that is it. I know that one possible initial trajectory is 70 degrees, but I need to find the other, and that is where I am having a problem. Of the equations that I know, it leaves me with two variables: theta and time.

My original idea was to take t=s_x/(25*sin(Θ)), and use that to replace the t's in s_y=V_0y*t + 1/2 a*t². The problem then becomes that I can't figure out how to solve for theta when I have cos and sin in the picture. (V_0y being replaced by 25*cos(Θ))

Is there an easy way to get this done, or can someone at least give me a push in the right direction? Thank you.

 

[Chestermiller]

What's the exact problem statement, please?

 

[Latsabb]

It is a multistep problem, and this is the last step, but I will write it out:

We throw a stone at a sharp upwards angle from the top of a steep cliff. The area below is flat, and lies 30 meters lower than the top of the cliff. The initial speed is 25 m/s and the initial trajectory is 70 degrees from the horizontal plane. We assume no air resistance.

a) How high over the starting point is the stone at it's highest point?
b) How long after being thrown does the stone hit the ground? Where does it hit?
c) Calculate the speed and angle when the stone hits the ground.
d) Is it possible to hit the same spot if we throw the stone with the same speed, but a different initial trajectory? If yes, calculate this angle.

So, I have already solved a, b and c, and I am now on d. The earlier calculations give me where on the x plane the stone will hit, which is 49.85 meters.

 

[Chestermiller]

OK. Much better. What's your final equation involving sinθ and cosθ?

 

[Latsabb]

(49.85*cos(θ)*sin(θ)-19.5)/sin^2(θ)+30=0

 

[Chestermiller]

(49.85*cos(θ)*sin(θ)-19.5)/sin^2(θ)+30=0

So, assuming you did the algebra correctly,
49.85\cotθ-19.5 \csc^2θ+30=0\csc^2θ=1+\cot^2θ49.85\cotθ-19.5 \cot^2θ+10.5=0

 

[Latsabb]

Hmmm... That would give that the angles are 20 and -80(ish) degrees, which cannot be correct. We don't use cot, csc in Norway, so maybe I misunderstood how those should have been converted over, but I will look over my algebra again, just in case.

 

[Chestermiller]

Hmmm... That would give that the angles are 20 and -80(ish) degrees, which cannot be correct. We don't use cot, csc in Norway, so maybe I misunderstood how those should have been converted over, but I will look over my algebra again, just in case.

Well, you could always substitute \phi = \frac{π}{2}-θ, and solve using tangent and secant. Also, check your equation by substituting θ = 70 degrees to see if it is satisfied.

 

[Latsabb]

I seem to have found my mistake. I reversed cos and sin when I was replacing t and the velocity, meaning the equation should have been:

(49.85*cos(θ)*sin(θ)-19.5)/cos^2(θ)+30=0

I have plugged in 70 for theta, and it works out. However, this could be rewritten as:

49.85*tan(θ)-19.5/cos^2(θ)+30=0

 

[Chestermiller]

I seem to have found my mistake. I reversed cos and sin when I was replacing t and the velocity, meaning the equation should have been:

(49.85*cos(θ)*sin(θ)-19.5)/cos^2(θ)+30=0

I have plugged in 70 for theta, and it works out. However, this could be rewritten as:

49.85*tan(θ)-19.5/cos^2(θ)+30=0

Then,49.85\tanθ-19.5\sec^2θ+30=049.85\tanθ-19.5\tan^2θ+10.5=0

I get -11.06°.

 

[Latsabb]

Yes, I had to look up how to manipulate a secant, but saw that as well. Thank you very much for the help.