Projectile hitting the incline plane horizontally

In summary, the projectile is thrown at angle θ with an inclined plane of inclination 45o. After time t, the velocity of the projectile is Vx=Velocity along the plane after time t and Vy=Velocity perpendicular to the plane after time t. The projectile eventually falls onto the plane again, and the angle θ is found to be tan-1(2) - 45o.
  • #1
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Homework Statement


A projectile is thrown at angle θ with an inclined plane of inclination 45o .
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]
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.

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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  • #2
Saurav7 said:
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).
 
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  • #3
tommyxu3 said:
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.
 
  • #4
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
tommyxu3 said:
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.
 
  • #6
Saurav7 said:
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.
 
  • #7
tommyxu3 said:
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. :)
 
  • #8
You can similarly discuss with everyone here like just now~ or maybe you can send me messages in the conversations directly also haha.
 
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1. What is a projectile?

A projectile is any object that is thrown or launched into the air and is subject to the force of gravity. This can include objects such as balls, rockets, or arrows.

2. How does a projectile hit an incline plane horizontally?

A projectile can hit an incline plane horizontally if it is launched at an angle that is parallel to the incline plane's surface. This means that the initial velocity of the projectile is perpendicular to the incline plane's surface.

3. What factors affect a projectile's trajectory when hitting an incline plane horizontally?

The factors that affect a projectile's trajectory when hitting an incline plane horizontally include the initial velocity, angle of launch, mass of the projectile, air resistance, and the angle and surface of the incline plane.

4. How does the angle of the incline plane affect the distance traveled by a projectile?

The angle of the incline plane can affect the distance traveled by a projectile by changing the amount of time the projectile spends in the air, and therefore, the distance it can cover before hitting the ground. A steeper incline plane will result in a shorter distance traveled, while a more gradual incline plane will result in a longer distance traveled.

5. What are some real-world applications of studying projectile motion on an incline plane?

Studying projectile motion on an incline plane can have many real-world applications, such as understanding the trajectory of a ball thrown during a baseball game, calculating the landing spot of a rocket launch, or predicting the distance a golf ball will travel on a sloped green. It is also important in engineering and designing structures such as ramps and ski slopes.

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