Projectile motion velocity homework

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To find the velocity of the projectile 1.50 seconds after firing, first calculate the vertical component of velocity using the formula Vy = Vy initial - g*t, where g is the acceleration due to gravity (9.8 m/s²). After determining the vertical velocity at 1.50 seconds, combine it with the constant horizontal velocity (Vx = 62 m/s) to find the total velocity. The horizontal component remains unchanged because there is no air resistance in the vacuum assumption. The final velocity can be expressed as a vector combining both components.
ruthi
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my problem says:
a projectile is fired with an initial speed of 75.2 m/s at an agle of 34.5 degrees above the horizontal on a long flat firing range. i figured out everyting so that

y initial = 0
y = 92.6m
Vy initial = 42.6 m/s
Vy = 0
g = 9.8 m/s^2
t = 8.69 s
Vx = 62 m/s

the question asks for the volocity of the projetile 1.50 s after firing. how do you go about figuring that out?
 
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Assuming a vacuum here, and a flat earth, the horizontal component remains constant, so you can ignore it for the moment, and concentrate on the vertical component of velocity and acceleration (which only has a vertical component anyway in this case). After you've done this, then re-include the horizontal component of velocity.
 
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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