# Isaac Physics Level 5 Throwing a Ball Up A Flat Slope QuestionWatch

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#1
This question has had me wracking my brain for a couple hours, here it is:
"What angle to the horizontal should you throw a ball up a slope of an angle 𝜃=10∘ to the horizontal in order for it to travel the furthest distance up the hill until its first bounce?"

I had initially thought that I should adjust for the slope by doing sin10 of the displacement, but because I want the furthest distance I also thought I could disregard the horizontal component of the velocity but it just leads me to getting theta as 10 (which definitely isn't the answer).

Anyone know how I should tackle a question like this? Thanks.
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1 month ago
#2
You have expressions for horizontal and vertical displacements; at the point at which the ball hits the slope, their ratio will be tan 10°. You can use this to solve for the relevant t in terms of 𝜃, and then use this value to get an expression for the horizontal displacement at that point, again in terms of 𝜃. Finally, find the value of 𝜃 that gives the maximum value of this h.d. by differentiating the expression w.r.t. 𝜃 and setting equal to zero as per usual.

Let me know how you get on?
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#3
(Original post by Justvisited)
You have expressions for horizontal and vertical displacements; at the point at which the ball hits the slope, their ratio will be tan 10°. You can use this to solve for the relevant t in terms of 𝜃, and then use this value to get an expression for the horizontal displacement at that point, again in terms of 𝜃. Finally, find the value of 𝜃 that gives the maximum value of this h.d. by differentiating the expression w.r.t. 𝜃 and setting equal to zero as per usual.

Let me know how you get on?
I'm not sure what you do with tan10, I first just did vcos(theta)t/vsin(theta)t = tan10 but that would mean theta = 10, what would I have do to get t in terms of theta?
Thanks.
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1 month ago
#4
(Original post by Yusuf_I)
I'm not sure what you do with tan10, I first just did vcos(theta)t/vsin(theta)t = tan10 but that would mean theta = 10, what would I have do to get t in terms of theta?
Thanks.
Are you throwing the ball in outer space? (You forgot your 0.5 g t²)
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#5
(Original post by Justvisited)
Are you throwing the ball in outer space? (You forgot your 0.5 g t²)
Thanks, I didn't notice that mistake. I got to a point where I could differentiate with respect to theta against the horizontal distance which left me with 1/5v^2(cos^2(theta)-sin^2(theta) +2cos(theta)sin(theta)tan(10)) which gives me theta as 5 which definitely isn't the answer.
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1 month ago
#6
I'll have a go at a complete solution myself, offline, to see where you might have gone wrong.

Is no one else interested in this question? Normally RDKGames is here in 30 seconds
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1 month ago
#7
Your big expression's bracket comes to cos 2𝜃 + sin 2𝜃 tan 10, and setting this to zero definitely doesn't give you 𝜃 = 5, try again
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#8
(Original post by Justvisited)
Your big expression's bracket comes to cos 2𝜃 + sin 2𝜃 tan 10, and setting this to zero definitely doesn't give you 𝜃 = 5, try again
I realised my mistake. I confused cotangent (cos/sin) and tangent (sin/cos) and I did the maths as if it was falling down the hill rather than being thrown up the hill (which gives 40 degrees), but I got the right answer in the end (50 degrees).

Thanks for the help!
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