# Projectile hitting the incline plane horizontally

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1. Jun 19, 2016

### Saurav7

1. The problem statement, all variables and given/known data
A projectile is thrown at angle θ with an inclined plane of inclination 45o .
Find θ if projectile strikes the inclined the plane horizontal

2. Relevant equations
Taking x-axis along the incline and y-axis perpendicular to incline.

Vx=ucosθ - gsint(45)t
Vy=usinθ - gcos(45)t

These are the velocities after time t.
Vx=Velocity along the plane after time t.
Vy=Velocity perpendicular to plane after time t.

3. The attempt at a solution
As at the time of horizontal collision with the incline. The projectile will make an angle of 45o 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 Vx=Vy
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=45o

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

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Last edited: Jun 19, 2016
2. Jun 19, 2016

### tommyxu3

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).

3. Jun 19, 2016

### Saurav7

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

4. Jun 19, 2016

### 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.

5. Jun 19, 2016

### Saurav7

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 dont know if this is the right result.

6. Jun 19, 2016

### tommyxu3

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.

7. Jun 19, 2016

### Saurav7

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. :)

8. Jun 19, 2016

### tommyxu3

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