Find the speed and direction of a particle which, when projected from a point 15m above the horizontal ground, just clears the top of a wall 26¼m high and 30m away.
sol: 25m/s and 36°52'
I have been at this for a while and cannot figure it. Not sure if I can assume that the maximum height is 26¼m or not, and even so, still can't figure out the equations and how to solve with 3 unknowns and only 2 equations
my starting point is this so far:
x = vt cos a and y = vt sin a - 4.9t² + 15
This is a problem of modeling ballistics motion.
When a object is fired at at the angle of at the height of with acceleration of gravity being
Its displacement is . Notice that I said displacement because this is a type of problem which is a vector-valued function. This give you the position at any giving time. The other formula in physics just gives you the height at any given time. Now you must agree that knowing the position at any given time would mean you know its height at any given time. Thus, this formula is superior to the other one.
Try doing your problem with this formula, if you cannot tell me and I will try to do it for you.
I am sure you are correct in what you say, but my course requirements don't actually get that far into physics, it is esentially a calculus chapter, with a dash of physics thrown in to make it difficult for us. I have zero physics knowledge and as far as i can tell in this chapter, all the questions are supposed to be answered from the basic two acceleration statements in parametric form
a(x) = 0 and a(y) = -9.8 (sorry for notation, that is supossed to be x with two dots above it and y with the same ... haven't progressed that far with my newly acquired, but severely minimal laTex skills yet)
from here, we integrate to get velocity statements
v(x) = vcosa and v(y) = vsina -9.8t where v = initial velocity and a = angle of projection
then from there we get
x = vt cos a and y = vt sin a - 4.9t² + 15 and usually manipulate that into a cartesian form with tan
i am sure there are probably much more sophisticated and easy methods to do this problem, but i have no idea about anything other than this method.
sorry for the trouble
I also have zero-physics knowledge I just use the fomulas.
You start at a height of 15 m. The ball travels 30 m and barely over the fence at 26 1/4m (which means its height was 26 1/4) at that time.
Thus at any given time, (using the formula I gave you before)
I agree with you you must need more. Perhaps the problem was saying at its maximum height?