Show that a projectile lands at a distance ##R = \frac{2v_0^2 sin \theta cos(\theta + \phi)}{g cos^2 \phi}##

In summary, @Redwaves solved the problem using the prescribed method and post #8 is not much simpler.
  • #1
Redwaves
134
7
Homework Statement
A ball is thrown with initial speed ##V_0## up an inclined plane. The plane is inclined at an angle ##\phi## above the horizontal, and the ball's initial velocity is at an angle ##\theta## above the plane. Choose axes with x measured up the slope, y normal to the slope.
Relevant Equations
##R = \frac{2v_0^2 sin \theta cos(\theta + \phi)}{g cos^2 \phi}##
##V_x = V_0 cos \theta ##
##x = V_0 cos \theta t##

##V_y = V_0 cos \theta ##
##y = V_0 cos \theta t##

##F_x = m\ddot{x}##
##-mgsin \phi = m\ddot{x}##
##\dot{x} = -gtsin\phi + V_x##
##x = -\frac{1}{2} gt^2 sin \phi + V_x t##
##x = -\frac{1}{2} gt^2 sin \phi + v_0 cos\theta t##

##F_y = m\ddot{y}##
##-mgcos \phi = m\ddot{y}##
##\dot{y} = -gtcos\phi + V_y##
##y = -\frac{1}{2} gt^2 cos \phi + V_y t##
##y = -\frac{1}{2} gt^2 cos \phi + v_0 sin \theta t##

I don't see where ##R## comes from.
 
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  • #2
You have some transcription errors in some equations (and what are x0, y0?), but you final equations are ok.
What is the value of y when it lands?
 
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  • #3
haruspex said:
You have some transcription errors in some equations (and what are x0, y0?), but you final equations are ok.
What is the value of y when it lands?
y = 0, since the ball is on the inclined plane?
 
  • #4
Redwaves said:
y = 0, since the ball is on the inclined plane?
Yes. So what do you deduce for x at that point?
 
  • #5
haruspex said:
Yes. So what do you deduce for x at that point?
I have to find t from the second equation then replace t in the x equation. Is it correct?
 
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  • #6
Redwaves said:
I have to find t from the second equation then replace t in the x equation. Is it correct?
Yes.
 
  • #7
I found it, thanks
 
  • #8
I suggest you ignore the bit about measuring x along the slope and y normal to the slope. In that case we have,
$$
v_{0x}=|v_0| \cos(\theta +\phi)
$$
$$
v_{0y}=|v_0| \sin(\theta +\phi)

$$
Neglecting wind resistance, the only force is in the y direction.
$$
ma_y=-mg
$$
$$
\frac{dv_y}{dt}=-g
$$
$$
y=v_{0y}t-\frac{1}{2}gt^2
$$
From ##\frac{dv_{0x}}{dt}=0##, the velocity in the x direction is constant, i.e. ##v_x=v_{0x}##.
$$
\frac{dx}{dt}=v_{0x}
$$
$$
t=\frac{x}{v_{0x}}
$$
Plugging the expression for ##t## into our equation for y we get
$$
y=\frac{v_{0y}}{v_{0x}}x-\frac{1}{2}g(\frac{x}{v_{0x}})^2
$$
The trajectory intersects the plane at
$$
y=R\sin(\phi)
$$
$$
x=R\cos(\phi)
$$
We thus have
$$
R\sin(\phi)=R\cos(\phi)\frac{v_{0y}}{v_{0x}}-\frac{1}{2}g\frac{R^2\cos^2(\phi)}{v_{0x}^2}
$$
Solving for ##R##,
$$
R=\frac{2|v|^2\cos(\theta + \phi)}{g\cos^2(\phi)}(\sin(\theta + \phi)\cos(\phi)-\cos(\theta + \phi)\sin(\phi))
$$
From trig addition angle formulas,
$$
\sin(\theta + \phi)=\sin(\theta)\cos(\phi)+\sin(\phi)\cos(\theta)
$$
$$
\cos(\theta + \phi)=\cos(\theta)\cos(\phi)-\sin(\phi)\sin(\theta)
$$
$$
R=\frac{2|v|^2\cos(\theta + \phi)}{g\cos^2(\phi)}(\sin(\theta)\cos^2(\phi)+\sin(\phi)\cos(\phi)\cos(\theta) -\sin(\phi)\cos(\phi)\cos(\theta) +\sin^2(\phi)\sin(\theta))
$$
$$
R=\frac{2|v|^2\cos(\theta + \phi)\sin(\theta)}{g\cos^2(\phi)}
$$
 
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  • #9
Jesus @Fred Wright , I haven't checked your solution in detail but it seems to me that you reveal the whole solution to the OP. Here in physics forums we are not allowed to do that. My only hope is that the OP is obliged to use x and y-axis as mentioned in the OP
 
  • #10
Delta2 said:
Jesus @Fred Wright , I haven't checked your solution in detail but it seems to me that you reveal the whole solution to the OP. Here in physics forums we are not allowed to do that. My only hope is that the OP is obliged to use x and y-axis as mentioned in the OP
My reading of post #7 is that, fortunately, @Redwaves solved it already using the prescribed method. And it doesn't seem to me that the method in post #8 is much simpler, if at all.
 
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Related to Show that a projectile lands at a distance ##R = \frac{2v_0^2 sin \theta cos(\theta + \phi)}{g cos^2 \phi}##

1. What is a projectile?

A projectile is any object that is thrown or launched into the air and moves under the force of gravity. Examples of projectiles include a baseball being thrown by a pitcher, a bullet fired from a gun, or a rocket launched into space.

2. What is the equation for the distance a projectile travels?

The equation for the distance a projectile travels is R = (v02sinθcos(θ+φ))/(gcos2φ), where v0 is the initial velocity, θ is the angle of launch, φ is the angle of the projectile's path with respect to the horizontal, and g is the acceleration due to gravity.

3. How is the distance a projectile travels affected by its initial velocity?

The distance a projectile travels is directly proportional to its initial velocity. This means that the greater the initial velocity, the farther the projectile will travel.

4. What role do angles play in the distance a projectile travels?

The angle of launch and the angle of the projectile's path with respect to the horizontal both play a role in determining the distance a projectile travels. The optimal angle of launch for maximum distance is 45 degrees, and any deviation from this angle will result in a shorter distance.

5. How does gravity affect the distance a projectile travels?

Gravity is the force that pulls a projectile towards the ground, causing it to follow a curved path. The greater the acceleration due to gravity, the shorter the distance a projectile will travel. This is why the distance a projectile travels is inversely proportional to the square of the cosine of the angle of launch.

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