# Mechanics 1 URGENT HELP needed

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#1
hello look at this question from the specimen paper
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#2
I am having major trouble understanding part 2, the way I see it its wrong ! I cant figure it out.

Shouldnt it say show that the vertical height of a droplet at time t is given by

y= 1 + 20tsina - 5t^2

NOT COS ??? cause cos if for the horizontal and we are only talking about the vertical in this question or am I just wrong 0
16 years ago
#3
Yes, it should be 20sin(alpha) for the vertical velocity component. It appears they've made a mistake. Not uncommon in specimen papers.

Regards,
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#4
That could explain it then, phew

I dont get what the hell part 3 is on about either or the rest of it Failure, it seems is inevitable for me
0
16 years ago
#5
Originally posted by mrmarsupial
That could explain it then, phew

I dont get what the hell part 3 is on about either or the rest of it Failure, it seems is inevitable for me
From now on a = alpha.

Well, we know that when the vertical height (y) is equal to zero, it has landed and thus reached its maximum horizontal range (X), thus, using the equation for the vertical height from part ii):

1 + 20tsin(a) - 5t^2 = 0, when it has landed.

Solving this quadratic yields two solutions for the time at which it lands:

5t^2 - 20tsin(a) - 1 = 0
(t - 2sin(a))^2 - 4sin^2(a) - 1/5 = 0
t = +/-Sqrt(4sin^2(a) + 1/5) + 2sin(a)

We can ignore the negative root as it would give us a negative time, which is obviously not relevant for this exercise.

We can now substitute this time value t into our equation from part ii). for the horizontal distance (x), which will then give us the maximum horizontal range (X) as we have set t equal to the time at which the water stream hits the ground:

X = 20tcos(a) = 20(Sqrt(4sin^2(a) + 1/5) + 2sin(a))cos(a)
= 20Sqrt(4sin^2(a) + 1/5)cos(a) + 40sin(a)cos(a)

We want to get rid of that surd (as there are none in their expression), so get that on its own on one side:

20Sqrt(4sin^2(a) + 1/5)cos(a) = X - 40sin(a)cos(a)
400(4sin^2(a) + 1/5)cos^2(a) = (X - 40sin(a)cos(a))^2
400(4sin^2(a) + 1/5)cos^2(a) = X^2 + 1600sin^2(a)cos^2(a) - 80Xsin(a)cos(a)

Cancelling the 1600sin^2(a)cos^2(a) term from both sides clears it up a bit:

80cos^2(a) = X^2 - 80Xsin(a)cos(a)
5X^2 - 400Xsin(a)cos(a) - 400cos^2(a) = 0

Multiply through by sec^2(a):

5sec^2(a)X^2 - 400Xtan(a) - 400 = 0

Remember Pythagoras (sec^2(A) = 1 + tan^2(A)), so:

5(1 + tan^2(a))X^2 - 400Xtan(a) - 400 = 0
5tan^2(a)X^2 - 400Xtan(a) + 5X^2 - 400 = 0

Hope that helps,

Regards,
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