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danni
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Homework Statement
A ball is thrown straight up. At the top of its path its instantaneous speed is
a) 5m/s
b) 20m/s
c) 50m/s
d) 10m/s
What is the Instantaneous Speed at the Top of a Ball's Path?
- Thread starter danni
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- Dimension Kinematics
AI Thread Summary
The instantaneous speed of a ball at the top of its path is zero, as it momentarily stops before descending. The question presents multiple choice options for speed, but none accurately reflect this fact. Understanding that at the peak of its trajectory, the ball's velocity is zero is crucial for solving similar problems. The lack of additional information like initial velocity or time complicates the solution process. Ultimately, recognizing that the instantaneous speed at the top of the path is zero is essential for answering such questions correctly.
Neglect gravity other than that of water; neglect friction.
F1=ρgsh=1000*10*1*(1+4)=50000 N;
F1=ρgsh=1000*10*0.1*4=4000 N;
F3=50000-4000=46000 N?
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}...
I was told forces are vectors so they can be divided into x and y components, but what about the normal force or the force of static friction? do you divide those into x and y components if say you had a block that was lying on an angled surface?
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