Catapult Projectile Question wow

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A catapult launches a rock from a height of 35.0 m at an angle of 48.7° with a speed of 26.1 m/s, and the goal is to calculate the horizontal distance traveled. The initial attempt at solving the problem involved using the quadratic formula but resulted in incorrect values due to a sign error in the equation. After correcting the equation to 4.9t² - 19.6t - 35 = 0, the user found the correct time of flight. This adjustment resolved the issue of obtaining imaginary answers, leading to a successful calculation of the projectile's distance. The discussion highlights the importance of careful equation setup in projectile motion problems.
dkbuono
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Homework Statement



Catapult
A catapult on a cliff launches a large round rock towards a ship on the ocean below. The rock leaves the catapult from a height H = 35.0 m above sea level, directed at an angle theta = 48.7° above the horizontal, and with a speed v = 26.1 m/s. Assuming that air friction can be neglected, calculate the horizontal distance D traveled by the projectile.


Homework Equations


y=y0+Vy0t-1/2gt^2
x=x0+ Vx0t



The Attempt at a Solution



Ok, So I plugged in my knowns to find the time for this, using the first equation setting it up like this 4.9t^2-19.6t+35, where vy0t=26.1*sin48.7 gave me 19.6 then I plugged in my values to the Quadratic formula where the answers I got were 2.97 and 1.03 respectively to find time, so I use the lower of the two numbers and I plug it back into x=x0+Vx0t, since intial D is 0 you add 26.1*cos(48.7) to give 17.2 then you times it to the time calculated 1.03 to get 17.72 m...its not right, I don't understand where I am going wrong I've set it up 7 different ways... This seems like a relatively easy question I only have 2 more tries left on it out of 10. Would someone point me in the right direction I am getting frustrated at this point and feel like and idiot considering its only ch.2.
 
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I would recalculate your quadratic
4.9t^2-19.6t+35

Only this time I would use 4.9t2 - 19.6t - 35 = 0

You might get a better result.
 
LowlyPion thank you so much you were right that simple error completely distorted everything! :)
 
dkbuono said:
LowlyPion thank you so much you were right that simple error completely distorted everything! :)

I got very suspicious when I saw the solution had 2 imaginary answers.

Glad it worked out.

Good Luck.
 
LowlyPion said:
I got very suspicious when I saw the solution had 2 imaginary answers.

I'm going to have to write that one down. :biggrin:
 
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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