Projectile Motion of golf chip shot

AI Thread Summary
To determine the initial speed required for a golf chip shot to reach a hole 60 m away at a launch angle of 50°, both horizontal and vertical motion equations must be used. The horizontal motion can be described by the equation xf = xi + vxi(t), while the vertical motion requires yf = yi + vyit + (1/2)at^2. The discussion emphasizes the need for an additional equation to account for vertical displacement. Participants seek clarification on how to apply these equations effectively. The conversation highlights the importance of integrating both components of motion to solve the problem accurately.
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Homework Statement



A golfer wishes to chip a shot into a hole 60 m away on flat level ground. If the ball sails off at 50°, what speed must it have initially? Ignore aerodynamic effects.

Homework Equations



xf = xi + vxi(t)
vx = vi(cos)angle


The Attempt at a Solution



How would I plug this in>
 
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How would I plug this in>
Into what?

You need one more equation. One that describes the vertical direction.
 
yf = yi + vyit + (1/2)at^2

this??
 
Yes, that one will work.
 
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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