me434343 Posted August 24, 2005 Posted August 24, 2005 hey i need help with a question, i need to be taught. So if you would please aswer this question for me and also tell me how you got that answer, that'd be great! a girl, standing on the roof of a house which is 5m high, throws a stone vertically upward. It strikes the ground 7s after being thrown. calculate: a) the initial velocity b) the maximum height the stone reaches (abouve the ground) c) the velocity of the stone as it strikes the ground
Daecon Posted August 24, 2005 Posted August 24, 2005 That depends on the force and speed in which she throws it upwards to start with, otherwise those answers could be anything.
Klaynos Posted August 24, 2005 Posted August 24, 2005 v= end velocity u= initial velocity a= acceleration (due to gravity) t= time s= distance from end point to start If you write that down and write down all that you know, then use the equations of motion to work out the ones you don't know. So for part a) you have: u=_ v=unknown s=-5m t=7s a=-9.81m/s/s So if you use: s=ut+1/2 ut2 You can find u, do the same for parts b, and c.
me434343 Posted August 24, 2005 Author Posted August 24, 2005 could i say that the distance from the roof (where it is thrown) back to the point of the roof is ther same... oh crap what about the person's hieght... that could effect something too... so confused
me434343 Posted August 24, 2005 Author Posted August 24, 2005 yeah.. maybe it's a question no one can answer.. a trick question.. damn i hate those..
Klaynos Posted August 24, 2005 Posted August 24, 2005 you should take the girls height into account, I didn't as we where given no info on it, I just too it as the ball was released 5m above the ground. You have no information about how high it gets, so for the initial velocity, you need to use the point where it hits the ground, and the time it takes to do that. After you have the initial velocity you can calculate the final velocity, and the maximam height (max height the final velocity v=0, you don't know the time, but you've already calculated u, and you are trying to work out the height so use: v2=u2+2as)
me434343 Posted August 24, 2005 Author Posted August 24, 2005 forget explaining it to me.. ill figure that out.. just tell me the answer
mezarashi Posted August 24, 2005 Posted August 24, 2005 The question doesn't concern the girl's height. Klaynos has already kindly written out the known and unknown parameters. 3 known and 2 unknown - which is solvable in this kind of kinematics problem. Simply plug the numbers into the equation. Additional info. The kinematics equations (a.k.a. the big four) that you should have memorized until you become familiar enough with these type of problems: http://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L6a.html
swansont Posted August 24, 2005 Posted August 24, 2005 In dealing with test questions that have ambguities like that, if you make any assumptions about information not given, state them clearly. Then, solve the problem, showing your work. In this case, you could assume zero height, and solve away. Or assume some height. If you use the proper equations and solve it correctly, you should get full credit. You hve to show your work, though, so a grader can see that you could have easily solved the problem if the assumtion had not been made or was different. An exception being if your assumption changes the nature of the problem, making it easier to solve. And if you ignored information that was given in favor of your assumption, all bets are off.
Daecon Posted August 24, 2005 Posted August 24, 2005 I think these questions are more fun: a girl, standing on the roof of a house which is 5m high, throws a stone vertically upward. It strikes her head 7s after being thrown.calculate: a) the initial velocity b) the maximum height the stone reaches (above the ground) c) the amount of pain inflicted and cost of medical treatment for concussion I know, I'm evil.
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