What is the time and speed of a rolling ball from tabletop to floor?

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To determine how long the ball is in the air, focus solely on the vertical motion, using the equation of motion for uniformly accelerated motion. The height of the tabletop (1.22 m) and the acceleration due to gravity will help calculate the time of flight. Once the time is established, the horizontal speed can be calculated by dividing the horizontal distance (1.58 m) by the time in the air. The horizontal motion is characterized by constant velocity, while vertical motion is influenced by gravity. This approach effectively separates the two components of motion for accurate results.
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Homework Statement


A small ball rolls horizontally off the edge of a tabletop that is 1.22 m high. It strikes the floor at a point 1.58 m horizontally away from the edge of the table. (a) How long is the ball in the air? (b) What is its speed at the instant it leaves the table?



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The Attempt at a Solution


So far all I've been able to calculate is the distance from the tabletop to the position on the floor that it landed. = 1.99m. Is there some equation with a trig function in it to get the time the ball is in the air?
 
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You have to split the motion into x and y components. The x motion is constant velocity and the y motion is uniformly accelerated downward. Stuff like y(t)=y0+v_y*t+(1/2)*a*t^2. Surely you know of stuff like that. No trig functions in this one.
 
For (a) all you need to know is how long the ball was in the air. Consider the ball along the y direction only. What is the inital velocity, displacement and acceleration? Stick thoose values into an equation of motion to find the value of time.

For (b) once you know the time the ball is in the air the rest should be easy. This time consider the ball only along the x axis. What would it's velocity have to be if it were to travel 1.58m in the time you calculated for (a).
 
Thanks, I do appreciate it.
 
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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