Further kinematics mechanics
Can I have help with Sliver part b) in the attached file?
(edited 3 years ago)
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I found the vertical speed using Pythagoras then I subbed it into the suvat formula v = u + at and i got 28.26m as the answer but the answers say it is 115m. What was your answer??
(edited 3 years ago)
Original post by sabreen1968
Can I have help with Sliver part b) in the attached file?
Are you able to find the times when the speed is 30 m/s?
Original post by sabreen1968
I found the vertical speed then i subbed it into the suvat formula v = u + at and i got 28.26m as the answers but i the answers say it is 115m. What was your answer??
Can you show your working?
Original post by sabreen1968
This is my working
Note there aee two values for a square root.
(edited 3 years ago)
Original post by mqb2766
Note there aee two values for a square root.
Why do we only use the negative value of the square root and not the positive? The negative value gives an answer of 115m which is correct. But the positive value does not give the correct answer.
(edited 3 years ago)
Original post by sabreen1968
Why do we only use the negative value of the square root and not the positive?
Just sketch the directions of motion on the graph. It must be heading downwards.
Im not sure the numbers make complete sense ... checking.
B must be above A?
(edited 3 years ago)
Original post by mqb2766
Just sketch the direction of motion on the graph.
Im not sure the numbers make complete sense ... checking.
B must be above A?
Im not sure the numbers make complete sense ... checking.
B must be above A?
Oh yes i understand. The question says that B is below A, so the speed is negative as the particle has to move down to reach B
Original post by sabreen1968
Oh yes i understand. The question says that B is below A, so the speed is negative as the particle has to move down to reach B
Is the book using g=10?
Original post by BuryMathsTutor
Is the book using g=10?
Yes
Original post by sabreen1968
Oh yes i understand. The question says that B is below A, so the speed is negative as the particle has to move down to reach B
Sort of, but the question must be wrong. Youre looking for the negative vertical speed solutiin, but B must be above A as 30 < initial speed.
Ok I have another question attached. Im confused with part b)
(edited 3 years ago)
Original post by mqb2766
What are the x-y coordinates when its due East?
y would be zero and x would have a value
Original post by mqb2766
can you solve then?
This is my working. I got stuck at solving the equation when i equaled y to 0 because it isnt a normal quadratic.
Original post by mqb2766
iits a hidden quadratic in sqrt(t).
Just use the quadratic equation if you can't spot the factors easily.
On the last line, its 2t^2
Just use the quadratic equation if you can't spot the factors easily.
On the last line, its 2t^2
Im not sure what you mean by on the last line it is 2t^2
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