Projectile motion of conservation of energy?

AI Thread Summary
To determine the initial speed required for a ball to pass through a hoop 20 meters away and 2 meters higher, the key is to use projectile motion equations. The angle of release is given as 45 degrees, which simplifies calculations. Time can be calculated using the height requirement and gravitational acceleration. The trajectory equation can be applied to find the initial velocity, with one participant suggesting a solution of 14.75 m/s. Understanding the derivation of motion equations is emphasized over memorization for effective problem-solving in physics.
firemanpato
Projectile motion of conservation of energy?

I have a problem that is stumping me. A guy throws a ball and wants it to go through a hoop 20 meters away and the hoop is 2 meters taller than from where he released the ball. If the only given info is the angle at which he threw the ball (45 degrees) How do I set up this problem to determine the initial speed at which the ball was thrown to make it go through the hoop. I have no idea how to set it up.:confused:
 
Physics news on Phys.org
Dang it! I just helped someone solve this exact problem a week ago.

I think the one thing that will help you the most is that the major problem here is that you really don't have the time or the initial velocity. With that in mind, one hint that should help is that you can determine the time using gravity and the height requirement!
 
how is this answer

:smile: since you know both the X and Y at the time the projectile pass the thread you can use the equation for the trijectory of the projectile. I solved it in that way and got answer as 14.75

y=Xtan(45)-(gX^2)/2(vcos45)^2

(refer any book for the equation i am not so sure about it but the value 14.75 have been tested and found correct.)
 
Although shanu's equation seems correct, memorizing loads of equations is hardly the way to do physics. The equation can be easily derived as we know that for constant acceleration:
x = x0 + v0xt + ½axt2
y = y0 + v0yt + ½ayt2

Two equations, two unknowns (as the equations simplify quite a bit with the information given in the problem).
 
Päällikkö said:
memorizing loads of equations is hardly the way to do physics
I agree 100%. In fact, I'd go one step further and say that it's even easier to start with F = ma and derive the 2 equations of motion of the projectile (but, only if you've had calculus). I had elementary physics x years ago (I won't tell you what x is :wink:), and had long since forgotten those 2 equations, but F=ma I'll always remember.:smile:
 
Last edited:
thank you for your advice. but it is the same i do. never memeorises the equation which i could get in a moment of calculation or thinking but when it is to take a long time i memorises them anlong with the way to derive it.

ANY WAY THANK U ALL FOR UR SUGESTIONS
 
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}...
Back
Top