Solve Easy Physics Question: Height of Freely Falling Object in 1.10s over 26.0m

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To solve the physics problem of determining the height from which a freely falling object descends, it is essential to recognize that the object takes 1.10 seconds to travel the last 26.0 meters before impact. The approach involves calculating the average speed during this final segment and relating it to the time taken to fall from the initial height. By denoting the time to fall the distance above the last 26 meters as T_1, the total fall time becomes T_1 + 1.10 seconds. The average speed can be calculated using both the given distance and time, and by applying kinematic equations. Equating the two expressions for average speed will allow for the determination of T_1 and ultimately the initial height.
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A certain freely failing object requires 1.10 s to travel the last 26.0 m before it hits the ground. From what height above the ground did it fall?

its a homework problem that i can't seem to crack, seems easy but i just started physics and I'm stuck, can someone give me a push in the right direction, maybe get me a basis on where to go with this problem and how to complete it? thanks,the help is greatly appreciated. :smile:

-ryan
 
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There are several ways to attack this problem. Don't be afraid to just jump in with whatever you know, even if you don't see how it will lead to an answer.

Here's one approach. Call the time it takes to fall from its starting point to 26m T_1. Now you know it takes a total time of T_1 + 1.10 s to complete the trip. Figure out the average speed it must be falling during that last 26m. Realize the the initial and final speeds (for that interval) are: Vi = g(T_1) & Vf = g(T_1 + 1.1).

Now it's your turn. Play around with this until you figure it out. (See if you can figure out T_1, and thus the total time it takes to fall.)
 
i must be a complete moron or not in the physics mode yet, but i still can't really figure out this problem. I'm probably getting stuck on something simple, but could someone give me another hint on how to get this problem done? its driving me nuts. :bugeye: thanks
 
Here's another hint: Find the average speed during that 26m distance. Do it two ways:
(1) Using the information given.
(2) Using the formulas I gave in terms of T_1.

Set those two equal and solve for T_1.
 
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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