How Do You Calculate the Range of a Projectile on a Sloped Surface?

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To calculate the range of a projectile on a sloped surface, the initial speed of the stone is 18 m/s, and the slope angle is 10 degrees. The correct approach involves setting up a coordinate system where the x-axis is parallel to the slope and the y-axis is perpendicular to it. The time of flight must be calculated considering the stone's vertical motion, which does not return to the same height due to the slope. The initial velocity vector should be adjusted to reflect the angle of the slope, leading to a more accurate calculation of the range. Ultimately, the solution requires careful consideration of the projectile's motion in relation to the slope's geometry.
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Homework Statement


A catapult projects a stone in the normal direction to a playground which slopes at a angle of 10 degrees to the horizontal.
The initial speed of the stone is 18 m/s. calculate the range parallel to the playground.

Homework Equations


the usual ones!

The Attempt at a Solution



My first equation is v = 18cos10i + 18sin10j
using v = u + at in the y direction with a = -9.8m/s^2 and v = 0 u =18sin10
using this i find that the time that the stone hits the ground is t = 0.32 seconds
now in the x direction x = vt since we ignore air resistance
i obtain 18cos10(0.32) = 5. 8 m

the answer in the back of the book is 11.8 m.

so i am wrong. i think i have interpreted the question incorrectly!
but i have tried setting the angle to 80 degrees, but that came out horribly wrong
 
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HMPARTICLE said:
My first equation is v = 18cos10i + 18sin10j
Where i is horizontal? No. A useful check is to consider what cos theta and sin theta give when theta = 0. In this case, that means a horizontal playground, so the stone goes straight up.
HMPARTICLE said:
using v = u + at in the y direction with a = -9.8m/s^2 and v = 0 u =18sin10
When considering the final speed of a projectile, you must take the speed just before it hits whatever it hits. The speed after it has landed is irrelevant.
 
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HMPARTICLE said:
My first equation is v = 18cos10i + 18sin10j
Does the initial velocity vector make a 10 degree angle to the horizontal?
using v = u + at in the y direction with a = -9.8m/s^2 and v = 0 u =18sin10
using this i find that the time that the stone hits the ground is t = 0.32 seconds

Looks like you are finding the time at which the vertical component of velocity is zero. Is this the time that you want to use for when the stone returns to the ground?

Also, note that the stone does not return to the same vertical height as it started due to the tilt of the playground.

I think things will work out more easily if you choose your coordinate system such that the x-axis runs down along the slope of the playground and the y-axis is perpendicular to the ground of the playground.
 

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if i was to use the suggested co-ordinate system wouldn't the cannon just fire at a 90 degree angle?
 
HMPARTICLE said:
if i was to use the suggested co-ordinate system wouldn't the cannon just fire at a 90 degree angle?

Yes, the cannon would fire at a 90 degree angle with respect to the x-axis running along the slope. If you do decide to choose your x-axis parallel to the slope and the y-axis perpendicular to the slope, then you will want to think about the x and y components of acceleration of the stone.
 
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DONE! Thank you so much guys!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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