Projectile Launching Question(multiple parts)

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The discussion revolves around a physics problem involving the launch of a 2.0 kg object from a 100 m cliff with an initial velocity of 40 m/s at a 30-degree angle. Participants are working through various parts of the problem, including expressing the initial velocity in "i,j" notation, calculating the maximum height reached by the object, and determining if it clears a 40 m obstacle located 200 m away. There is a focus on calculating the time it takes for the object to reach the barrier and the relevance of horizontal velocity in solving part C. The need for clarity in calculations and the importance of showing work to identify errors is emphasized throughout the discussion.
brayr0019
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Heres the Problem...
A 2.0 kg object is launched from a 100 m tall cliff with an initial velocity of Vo=40m/s @ 30 degrees. There is a 40 m tall obstacle 200m from the base of the cliff.
a)write initial velocity in "i,j" notation
-I found this to be 34.64m/s i + 20m/s j = Vo
b) how far above the launch point does the ball travel?
- I also found this... h= 20.41m
c) prove that the object makes it over the barrier, i.e. what is the height of the object at x=200m?
d) how far from the base of the cliff does the object land?
e) what is the velocity of the object as it hits the ground(write velocity in "i,j" notation and as a magnitude and angle)


Any help at all for parts C,D, and/or E would be much appreciated! thanks everyone!
 
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Forum rules require you to have a go and show your working so we can see where you are going wrong.
 
Heres my work. Clearly my numbers are wrong and I don't know where I went wrong so I can't even go about trying part E until i finish that.
 

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Ok looking at your answer for C...

It looks like you have started by calculating the time (2.04 seconds) taken to reach the top of the trajectory. I'm not sure that's relevant to part C. The problem asks for the height at distance 200m. If you want to work out the time when the object reaches the barrier (200m) think about the horizontal velocity.
 
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