View Full Version : Equation problem.
Gareth56
16th November 2007, 08:51 PM
Can someone please explain how the answer h is arrived at because it appears on both sides of the equation?
t1 = 0.69s & t2 = 2.37s
h =((h+0.5gt1^2)t1)t2 - 0.5gt2^2
The answer is h = 8.02m
I tried attaching small scan of the equation from the book but despite the file size only being 12KB the system rejected it :(
KatieB
16th November 2007, 09:11 PM
What was the book and page number. I may have it?
I find it difficult to understand your equation but normally you would bring terms containing the subject h to one side of the equation.
factorise that side contining h
i.e. h(........)=b say
so h= b/(........)
Hope that helps
Gareth56
16th November 2007, 10:16 PM
I tried to attach a 12KB file to make it easier to see the equation but I got the message "UPLOAD ERROR" that it exceed my quota by 8.4 KB. Does anyone know what this means because according to the instructions I can attach a JPG file up to ~39KB
Does this make any more sense?
t1 = 0.69s & t2 = 2.37s g =9.81m/s
h =(h+0.5gt1^2)/t1)t2 - 0.5gt2^2
The answer is h = 8.02m
talkingturtles
20th November 2007, 08:54 PM
t1 = 0.69s & t2 = 2.37s g =9.81m/s
h =(h+0.5gt1^2)/t1)t2 - 0.5gt2^2
Firstly put the values you have for t1, t2, and g into your eqution:
h=((h+ 0.5x9.81x0.69)/0.69)x2.37 - 0.5x9.81x2.37
Expand the sets of brackets:
h=(h/0.69 + 0.5x9.81x0.69/0.69)x2.37 - 0.5x9.81x2.37
h=(2.37h)/0.69 + 0.5x9.81x0.69x2.37 - 0.5x9.81x2.37
Put the numbers into a calculator:
h=(2.37h)/0.69 + 8.0211465 - 27.5508945
h=(2.37h)/0.69 - 19.529748
multiply by 0.69:
0.69h=2.37h - 13.47552612
subtract 0.69h from each side:
0 = 1.68h - 13.47552612
divide by 1.68:
0 = h - 8.0211465
subtract 8.0211465 from each side:
h=8.02 (3sig fig)
and theres your answer!
Hopethat helps you work through it. <3
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