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#1
Question: 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?
Source:https://isaacphysics.org/questions/t...4-f302d846f53c

My working so far:
Since its being thrown up a slope, I resolved g into its perpendicular (gcos10) and parallel (gsin10) components. I then used SUVAT for the components parallel:
u=Vsin(x-10)
v=0
a=-gcos10
T=t

x is the angle the ball must be thrown at to the horizontal so (x-10) is the angle between the slope and its launch.

To find t, I used v=u+at which gave t=(Vsin(x-10))/gcos10

T= time of flight
so as T=2t , T=(2vsin(x-10))/gcos10

The horizontal velocity I would've thought would need to take into account the parallel acceleration so I'm stuck on finding an equation for the parallel velocity component which I think I can then use to find the angle..?
0
4 years ago
#2
Hi there,

While you're waiting for an answer, did you know we have 300,000 study resources that could answer your question in TSR's Learn together section?

We have everything from Teacher Marked Essays to Mindmaps and Quizzes to help you with your work. Take a look around.

If you're stuck on how to get started, try creating some resources. It's free to do and can help breakdown tough topics into manageable chunks. Get creating now.

Thanks!

Not sure what all of this is about? Head here to find out more.
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#3
bump
0
4 years ago
#4
(Original post by BP_Tranquility)
Question: 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?
Source:https://isaacphysics.org/questions/t...4-f302d846f53c

My working so far:
Since its being thrown up a slope, I resolved g into its perpendicular (gcos10) and parallel (gsin10) components. I then used SUVAT for the components parallel:
u=Vsin(x-10)
v=0
a=-gcos10
T=t

x is the angle the ball must be thrown at to the horizontal so (x-10) is the angle between the slope and its launch.

To find t, I used v=u+at which gave t=(Vsin(x-10))/gcos10

T= time of flight
so as T=2t , T=(2vsin(x-10))/gcos10

The horizontal velocity I would've thought would need to take into account the parallel acceleration so I'm stuck on finding an equation for the parallel velocity component which I think I can then use to find the angle..?
The acceleration parallel to the slope will be g sin10 down the slope.
0
4 years ago
#5
(Original post by TSR Learn Together)
Hi there,

While you're waiting for an answer, did you know we have 300,000 study resources that could answer your question in TSR's Learn together section?

We have everything from Teacher Marked Essays to Mindmaps and Quizzes to help you with your work. Take a look around.

If you're stuck on how to get started, try creating some resources. It's free to do and can help breakdown tough topics into manageable chunks. Get creating now.

Thanks!

Not sure what all of this is about? Head here to find out more.

It is very lengthy to explain on line the process especially since I do not know your level.

However I am attaching 2 questions of very similar nature that might help answer your particular problem.

PDF.pdf.

I hope they make sense.
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