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

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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.

"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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#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?

Let me know how you get on?

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(Original post by

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?

**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?

Thanks.

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#4

(Original post by

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.

**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.

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(Original post by

Are you throwing the ball in outer space? (You forgot your 0.5 g t²)

**Justvisited**)Are you throwing the ball in outer space? (You forgot your 0.5 g t²)

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#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

Is no one else interested in this question? Normally RDKGames is here in 30 seconds

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#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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(Original post by

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

**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

Thanks for the help!

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