Projectile hitting the incline plane horizontally

[Saurav7]

Homework Statement

A projectile is thrown at angle θ with an inclined plane of inclination 45^o .
Find θ if projectile strikes the inclined the plane horizontal

Homework Equations

Taking x-axis along the incline and y-axis perpendicular to incline.[/B]
V_x=ucosθ - gsint(45)t
V_y=usinθ - gcos(45)t

These are the velocities after time t.
V_x=Velocity along the plane after time t.
V_y=Velocity perpendicular to plane after time t.

The Attempt at a Solution

As at the time of horizontal collision with the incline. The projectile will make an angle of 45^o with the incline and hence the velocity of projectile V=Vcos45i + Vsin45j which means the x and y components of velocity are equal (as sin45=cos45).
And hence I applied the condition that V_x=V_y
that gives us:
ucosθ - gsint(45)t = usinθ - gcos(45)t
=> ucosθ-usinθ = gsin(45)t-gcos(45)t
=>u(cosθ-sinθ) = gt(sin45-cos45)
=>u(cosθ-sinθ)=0 (as sin45=cos45)
=>cosθ-sinθ=0
=>cosθ=sinθ => Tanθ=1
which gives θ=45^o

But the answer is θ = tan^-1(2) - 45^o
I know how the right solution came but I am more bothered about what was wrong in my attempt that I got it incorrect.

 

[tommyxu3]

Vx=Vy

This is right, but mint the same quality is their "magnitude" so you must put $|u\cos\theta - g\sin\frac{\pi}{2}t |= |u\sin\theta - g\cos\frac{\pi}{2}t|.$ Then the answer is one of the condition (throwing upwards).

 

[Saurav7]

This is right, but mint the same quality is their "magnitude" so you must put $|u\cos\theta - g\sin\frac{\pi}{2}t |= |u\sin\theta - g\cos\frac{\pi}{2}t|.$ Then the answer is one of the condition (throwing upwards).

I can't solve it further.Can you help me a bit further?
Thanks in advance.

 

[tommyxu3]

You will get $u(\cos\theta+\sin\theta)=\sqrt{2}gt,$ (of course the other direction of your original one), but you cannot solve it with just this. My hint is to observe that the object finally falls on the plane again.

 

[Saurav7]

You will get $u(\cos\theta+\sin\theta)=\sqrt{2}gt,$ (of course the other direction of your original one), but you cannot solve it with just this. My hint is to observe that the object finally falls on the plane again.

Thanks a lot brother. I got u(cosθ+sinθ)=√2gt and this happens at t=Time of flight so we get
t=T=2usinθ/(gcos45)
Putting in the values I got 3sinθ+cosθ=0
Then dividing by cosθ on both sides,we get;
3tanθ+1=0
and finally θ=arctan(-1/3)

I don't know if this is the right result.

 

[tommyxu3]

Putting in the values I got 3sinθ+cosθ=0

Plugging the value you may get $u(\cos\theta+\sin\theta)=\sqrt{2}gt=4u\sin\theta,$ so it must be $\cos\theta=3\sin\theta$ instead of what you get.
You can get $\arctan \frac{1}{3}=\arctan{2}-\frac{\pi}{4}$ by some trigonometric knowledge.

 

[Saurav7]

Plugging the value you may get $u(\cos\theta+\sin\theta)=\sqrt{2}gt=4u\sin\theta,$ so it must be $\cos\theta=3\sin\theta$ instead of what you get.
You can get $\arctan \frac{1}{3}=\arctan{2}-\frac{\pi}{4}$ by some trigonometric knowledge.

Man,you are a life saver to me. I was stuck on this since yesterday night :D . Thanks a lot,a lot. :D
Is there any other way to connect with you in case I get stuck again to some other question :P
Like can I PM you or something :D
Again,Thanks a lot. :)

 

[tommyxu3]

You can similarly discuss with everyone here like just now~ or maybe you can send me messages in the conversations directly also haha.