Ball Thrown Up An Inclined Plane

A yellow ball is thrown with initial speed v0=2v_{0} = 2 m/sm/s up an inclined plane. The plane is inclined at an angle 1515 ^ \circ above the horizontal, and the ball's thrown with an angle 4545 ^ \circ above the plane. If RR denotes the ball land's distance, find the value of RR. (assume g=10g = 10 m/s2m/s^2)

I found that the answer is 25(31)\frac{2}{5}(\sqrt{3} - 1) mm, but I'm not sure. Could anyone clarify? Thanks :)

Note by Nabila Nida Rafida
5 years, 11 months ago

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vx=2cos60=1v_{x} = 2 \cos 60^\circ = 1

vy=2sin60=3v_{y} = 2 \sin 60^\circ = \sqrt{3}

x=vxt=tx = v_{x}t = t

y=vyt12gt2=3t5t2y = v_{y}t - \frac{1}{2}gt^2 = \sqrt{3}t - 5t^2

Because tan15=yx,\tan 15^\circ = \frac{y}{x}, then x=t=y(2+3)x = t = y(2 + \sqrt{3})

By substituting t=y(2+3)t = y(2 + \sqrt{3}) to y=3t5t2,y = \sqrt{3}t - 5t^2,

y=3(2+3)y5(2+3)2y2y = \sqrt{3}(2 + \sqrt{3})y - 5(2 + \sqrt{3})^2y^2

y(5(7+43)y2(1+3))=0y(5(7 + 4\sqrt{3})y - 2(1 + \sqrt{3})) = 0

Since y0,y \neq 0,

y=2(1+3)5(7+43)y = \frac{2(1+\sqrt{3})}{5(7+4\sqrt{3})}

Because sin15=yR,\sin 15^\circ = \frac{y}{R}, then R=ysin15R = \frac{y}{\sin 15^\circ}

Hence, R=42(23)5R = \frac{4\sqrt{2}(2-\sqrt{3})}{5}

Fariz Azmi Pratama - 5 years, 11 months ago

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Firstly, good question Nabila and great effort on your answer.

Fariz nice approach to the question, I completely agree with your answer as I verified it using a different method; which was finding the intercepts of the trajectory of the projectile and the plane.

I found using simple physics formulae that the height is 0.15m0.15m

I then used this to model a formula for the trajectory; y=ax2+bxy=ax^{2}+bx where b is the initial gradient of the projectile tan(π/3)tan(\pi/3),

Using height of 0.150.15, aa was found to be 5-5

Hence, y=3x5x2y=\sqrt{3}x-5x^{2} is the trajectory in the Cartesian coordinate system

This is now intercepted with the plane which is linear; y=bxy=bx, were b is tan(π/12)tan(\pi/12)

Therefore the plane is represented as; y=(23)xy=(2-\sqrt{3})x

Furthermore, the coordinates of the intercepts of the trajectory and the plane are calculated to be:

y=65(32)y=\frac{6}{5}(\sqrt{3}-2) and x=25(31)x=\frac{2}{5}(\sqrt{3}-1)

Finally, using Pythagoras's Theorem, R=(25(31))2+(65(32))2R=\sqrt{(\frac{2}{5}(\sqrt{3}-1))^{2}+(\frac{6}{5}(\sqrt{3}-2))^{2} }

Hence, R=452(23)R=\frac{4}{5}\sqrt{2}(2-\sqrt{3})

Joel Jablonski - 5 years, 11 months ago

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Nice one!

Nabila Nida Rafida - 5 years, 11 months ago

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Mathematical approach, cool :)

Fariz Azmi Pratama - 5 years, 11 months ago

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Thanks, it seems my answer yet completed

Nabila Nida Rafida - 5 years, 11 months ago

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Nice solution, though just a slight mistake in the end. I believe, you can easily get the right answer if you try to deduce a point in the trajectory where yx=tan15 \frac {y} {x} = \tan 15^\circ using the above equations. This would give the required R most probably.

Lokesh Sharma - 5 years, 11 months ago

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Would you like to fix it, Lokesh? I'm not sure about my answer.....

Fariz Azmi Pratama - 5 years, 11 months ago

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@Fariz Azmi Pratama Referring to your solution, equation (3) and equation (4) define the trajectory of ball. The ball lands at the point (x,y) (x, y) such that yx=tan15 \frac {y} {x} = tan 15^ \circ .

Solving these three equations, the point can be located and hence the distance where the ball lands. Diving equation (4) with equation (3), yx=35t \frac {y} {x} = \sqrt3 - 5t and using yx=tan15 \frac {y} {x} = tan 15^ \circ , finally 35t=tan1535t=2(3) \sqrt3 - 5t = tan 15^ \circ \Rightarrow \sqrt3 - 5t = 2 - \sqrt(3) .

Solving this we have x=t=2325x = t =\frac {2 \sqrt3 - 2}{5} . Using trigonometry we get xR=cos15\frac{x} {R} = cos 15^ \circ and hence R=xcos15R=4(232)5(6+2)R = \frac{ x }{ cos 15^ \circ} \Rightarrow R = \frac {4(2 \sqrt3 - 2)}{5(\sqrt6 + \sqrt2)} .

Finally, R=4(226)5 R = \frac {4(2\sqrt2 - \sqrt 6)}{5} which is same as yours. Hahaha. I was wrong about your solution. Sorry about that one.

Lokesh Sharma - 5 years, 11 months ago

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@Lokesh Sharma Haha it's okay.

Fariz Azmi Pratama - 5 years, 11 months ago

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Range=2sin455cos215Range=\frac{ 2*\sin45 }{ 5 *\cos^215 }

Cody Martin - 5 years, 11 months ago

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it should be 4 (sqrt2)/5

Viji Raj - 5 years, 11 months ago

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Yes Victor is right. You may even use trigonometry sin 15 = (sqrt3-1)/2

Viji Raj - 5 years, 11 months ago

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Hey, I think what you found was incorrect in the sense that what you found wasn't what the question was asking for. I found that your answer was what I got for the horizontal distance, not the actually length between the ball land's distance. What I did was find the equation for both lines (the inclined plane and the trajectory) find the intersection which is where you got, and the last bit is apply the Pythagorean theorem to get that length.

Victor Phan - 5 years, 11 months ago

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45(2632)\frac{ 4 }{ 5 }(2\sqrt{6}-3\sqrt{2}) ??

Cody Martin - 5 years, 11 months ago

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How did you get the answer?

Fariz Azmi Pratama - 5 years, 11 months ago

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yes.I agree

David L. - 5 years, 11 months ago

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tan15=y/x =(2.sin60.t-1/2.10. t^2)/(2.cos60.t)
2-sqrt{3} =sqrt{3}-5t
t={2/5}(sqrt{3}-1)
R= x/sin 15 =t/sin 15 = {4/5}sqrt{2}

Viji Raj - 5 years, 11 months ago

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R = x / cos 15

Lokesh Sharma - 5 years, 11 months ago

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Yes you are right Lokesh. R=x/cos15 that gives me the same answer as yours

Viji Raj - 5 years, 11 months ago

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rotate Oxy 15

gx= -10sin15; gy= -10cos15

t = 2vsin45/gy

R=vcos45t - 1/2 * gx * t^2 =2*(1 - tan15)/cos15

thanh nguyen - 5 years, 10 months ago

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We need to start by finding final speed, vfv_f.

After that, u can start using conservation of mechanical energy

Ki+Ui=Kf+UfK_i + U_i = K_f + U_f

12mv02=12mvf2+mgRsin45\frac{1}{2}mv_0{}^{2} = \frac{1}{2} mv_f{}^{2} + mgR\sin 45^{\circ}

v02=vf2+2gRsin45v_0{}^{2} = v_f{}^{2} + 2gR\sin 45^{\circ}

Now u have all the values except RR and u can easily find RR

Saad Haider - 5 years, 10 months ago

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What is the value of vfv_f?

Lokesh Sharma - 5 years, 10 months ago

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U can use the equation of kinematics

In the y-direction

vyf2=(v0sin45)22gRsin45v_{yf}^{2} = (v_0\sin45^{\circ})^{2} -2gR\sin45^{\circ}

vyf=v0sin45gtv_{yf} = v_0\sin 45^{\circ} - gt

Rsin45=v0sin45t12gt2R\sin 45^{\circ} = v_0\sin 45^{\circ}t - \frac{1}{2} gt^{2}

Rsin45=12(vyf+v0sin45)tR\sin 45^{\circ} = \frac{1}{2}(v_{yf} + v_0\sin 45^{\circ})t

In the x-direction

Rcos45=v0cos45tR\cos 45^{\circ} = v_0\cos 45^{\circ}t

Using the equations, we can find vyfv_{yf}. We also have vxf=vxi=v0cos45v_{xf} = v_{xi} = v_0\cos 45^{\circ}. Then u can find vfv_f by Pythagoras theorem: vf=vxf2+vyf2v_f = \sqrt{v_{xf}^{2}+v_{yf}^{2}}

Saad Haider - 5 years, 10 months ago

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i think it would be R=uusin2*/g=0.4m

abhimanyu manu - 5 years, 11 months ago

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