Projectile Motion Time Calculation

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An object fired up a frictionless ramp at a 65-degree angle with an initial velocity of 35 m/s takes approximately 7.88 seconds to return to its starting point, correcting an earlier calculation of 6.5 seconds. The key to solving this problem involves using a coordinate system aligned with the ramp and applying the equations of motion, specifically F=ma, to determine the acceleration due to gravity along the incline. By calculating the time using the formula t = 70/gsin65, the correct time is derived. The discussion emphasizes the importance of proper diagramming and component analysis in projectile motion problems. Ultimately, understanding the forces at play leads to accurate time calculations in projectile motion scenarios.
Destrio
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Hey,
An object is fired up a frictionless ramp 65 degrees to the horizontal with an initial velocity of 35 m/s, how long does the object take to return to the starting point.

What I did was:

cos 23 * 35m/s = Vyo
Vyo = 31.72 m/s

Vf = Vo + at
0 m/s = 31/72 m/s + (9.8m/s^2)(t)
t = 3.24s

Then doubled time for down to equal: 6.5s

The answer in the key is 7.9s.
What did I do wrong here?

Thanks,
 
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First thing I would do is draw a diagram. I would define a coordinate system in which the x-axis is parallel to the surface of the plane. That way all my calculations will deal with breaking components into the x-direction. Next I would use F=ma to calculate the acceleration of the object. Then apply projectile motion equations to calculate the time.
 
I'm still getting 6.5s as an answer :/
 
Starting with F=ma. The only force that is going to cause an acceleration is gravity (if we use the coordinate system I described).

F = ma
mgsin65 = ma
a = gsin65

Vf = Vi + at
0 = 35 m/s + (gsin65)t
t = 35/gsin65
multiply it by 2 to get whole time
t = 70/gsin65
t = 7.88 seconds.
 
ah,
that makes sense now
thanks very much
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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