How Do You Calculate Final Y-Velocity of a Projectile?

AI Thread Summary
To calculate the final y-velocity of a projectile launched at 40 m/s at a 38-degree angle, the x-velocity remains constant at 31.52 m/s due to ideal conditions. The projectile's angle of trajectory changes to 24 degrees below the horizontal, which affects the y-velocity. Using trigonometric functions, the final y-velocity can be determined by applying the tangent of the angle to the x-velocity. This results in a final y-velocity of approximately 14.03 m/s directed downward. The calculations confirm the relationship between the angles and the velocities in projectile motion.
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Homework Statement


A projectile is launched at 40 m/s and an angle of 38 degrees. After some time, its angle of trajectory is 24 degrees below the horizontal. What is the final velocity in the y direction at this time? Answer in m/s.



Homework Equations

 
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Find the x-axis motion... so... cos 38 * 40 = 31.52 m/s.. now that never changes in ideal conditions...only in the y direction the velocity changes due to gravity.. now u no at some time its 24 degrees below horizontal.. the ball or what ever is still traveling at 31.52 metres/s in the x direction.. and you have and angle.. so using triangle formulas... tan 24 * 31.52m/s = 14.0333m/s towards the ground.. i think this is right.
 
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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