Projectile Motion Kenimatics in Two Deminisions?

AI Thread Summary
To find the maximum height of a projectile launched at an initial velocity of 4.736 m/s and an angle of 40 degrees, the correct formula involves squaring the vertical component of the initial velocity, which is derived from the sine of the angle. The formula used should be (sin^2(40) * Vo^2) / (2 * g), where g is the acceleration due to gravity at -9.8 m/s². Concerns were raised about the initial calculations yielding a low height, indicating a potential misunderstanding of the formula. The final consensus confirms that using the squared sine function with the initial velocity squared, divided by twice the gravitational acceleration, is indeed the correct approach. This method will yield the accurate maximum height for the projectile's motion.
alexas
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Homework Statement


Find the maximum height of a projectile that as the following velocity and is launched at that angle.
Intial Velocity: 4.736 m/s
Angle (in degrees): 40
Gravity: -9.8 ?


I was thinking...

{((-4.736m/s)^2)(sin(40))}/{(-2)(g)} but the hieght comes out extremely low...

I have a feeling i am using a wrong formula... Any ideas?
 
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You really should be squaring the (sin40*Vo) but otherwise it won't go all that high will it, if the vertical component of Vo is about 1/3 of what gravity will knock down in 1 sec?
 
Sorry I meant to say the sin40 was also squared. Is the answer I get for thre sin^2 times v^2 divided by 2g the correct one?
 
alexas said:
Sorry I meant to say the sin40 was also squared. Is the answer I get for thre sin^2 times v^2 divided by 2g the correct one?

Looks ok to me.
 
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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