A yellow ball is thrown with initial speed \(v_{0} = 2\) \(m/s\) up an inclined plane. The plane is inclined at an angle \(15 ^ \circ\) above the horizontal, and the ball's thrown with an angle \(45 ^ \circ\) above the plane. If \(R\) denotes the ball land's distance, find the value of \(R\). (assume \(g = 10\) \(m/s^2\))

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

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TopNewest\[v_{x} = 2 \cos 60^\circ = 1\]

\[v_{y} = 2 \sin 60^\circ = \sqrt{3}\]

\[x = v_{x}t = t\]

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

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

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

\[y = \sqrt{3}(2 + \sqrt{3})y - 5(2 + \sqrt{3})^2y^2\]

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

Since \(y \neq 0,\)

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

Because \(\sin 15^\circ = \frac{y}{R},\) then \(R = \frac{y}{\sin 15^\circ}\)

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

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Firstly, good question

Nabilaand great effort on your answer.Fariznice 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 \(0.15m\)

I then used this to model a formula for the trajectory; \(y=ax^{2}+bx\) where

bis the initial gradient of the projectile \(tan(\pi/3)\),Using height of \(0.15\), \(a\) was found to be \(-5\)

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

This is now intercepted with the plane which is linear; \(y=bx\), were

bis \(tan(\pi/12)\)Therefore the plane is represented as; \(y=(2-\sqrt{3})x\)

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

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

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

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

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Nice one!

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Mathematical approach, cool :)

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Thanks, it seems my answer yet completed

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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 \( \frac {y} {x} = \tan 15^\circ \) using the above equations. This would give the required R most probably.

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

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

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

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

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

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We need to start by finding final speed, \(v_f\).

After that, u can start using conservation of mechanical energy

\(K_i + U_i = K_f + U_f\)

\(\frac{1}{2}mv_0{}^{2} = \frac{1}{2} mv_f{}^{2} + mgR\sin 45^{\circ}\)

\(v_0{}^{2} = v_f{}^{2} + 2gR\sin 45^{\circ}\)

Now u have all the values except \(R\) and u can easily find \(R\)

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What is the value of \(v_f\)?

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U can use the equation of kinematics

In the y-direction

\(v_{yf}^{2} = (v_0\sin45^{\circ})^{2} -2gR\sin45^{\circ}\)

\(v_{yf} = v_0\sin 45^{\circ} - gt\)

\(R\sin 45^{\circ} = v_0\sin 45^{\circ}t - \frac{1}{2} gt^{2}\)

\(R\sin 45^{\circ} = \frac{1}{2}(v_{yf} + v_0\sin 45^{\circ})t\)

In the x-direction

\(R\cos 45^{\circ} = v_0\cos 45^{\circ}t\)

Using the equations, we can find \(v_{yf}\). We also have \(v_{xf} = v_{xi} = v_0\cos 45^{\circ}\). Then u can find \(v_f\) by Pythagoras theorem: \(v_f = \sqrt{v_{xf}^{2}+v_{yf}^{2}}\)

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rotate Oxy 15

gx= -10

sin15; gy= -10cos15t = 2

vsin45/gyR=v

cos45t - 1/2 * gx * t^2 =2*(1 - tan15)/cos15Log in to reply

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}

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R = x / cos 15

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

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\[\frac{ 4 }{ 5 }(2\sqrt{6}-3\sqrt{2})\] ??

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How did you get the answer?

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yes.I agree

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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.

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it should be 4 (sqrt2)/5

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

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

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