A somewhat tricky projectile problem

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A spring-loaded gun can fire a projectile to a maximum height h when shot straight up. When fired at a 45-degree angle from the vertical, the maximum height reached is h/2. The user attempted to solve the problem using kinematics but struggled with the calculations and understanding the concepts without numerical values. They expressed a need for tips on improving problem-solving skills in physics. The discussion emphasizes the importance of understanding projectile motion and kinematic equations in solving such problems.
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Homework Statement



A sping-loaded gun can firen a projectile to a height h if it is fired straight up. If the same gun is pointed at an angle of 45degrees from the vertical, what maximum height can now be reached by the projectile?





The Attempt at a Solution



The answer is h/2. I don't know how to get there. I tried by using one of the kinematics equations but it didn't work out. This is my attempt:

vy = vsin(x)t + .5gt2

vy = v/sqrt(2)



This is what I tried to do. It's obviously wrong and I got stuck at that part. I'm not used to doing problems without numbers. If any of you have any tips on how to get better at them please share your tips.
 
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Well at the max height, the final vertical velocity is zero.

so if

v_y^2=u^y^2+2gs

what would be the height,s?
 
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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