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M1 Projectile Question - Confused
Hi,
Spent an age on this question and im still confused
The question is:
A golfball is driven from the tee with speed 30√2 meters per second at an angle α to the horizontal.
i) Show that during its flight the horizontal and vertical displacments x and y of the ball from the tee satisfy the equation:
y = x tanα - (x^2/360)(1+ tan^2 α
DONE
ii) The golf ball just clears a tree 5m high which is 150m horizontally from the tee. Find the two possible values of tanα.
DONE - ANSWERS 1.8 AND 0.6
iii) Use the discriminant of the quadratic equation in tanα to find the greatest distance by which the golf ball can clear the tree and find the value of tanα in this case.
Confused - Ok so i can understand that b^2 - 4AC must = 0 for the highest point. But how do you answer it? Its most probs very easy but my excuse is im tired lol.
Many thanks!
BTW The question is in MEI Structured Mathmatics, M1 Edition 3. Excersise 6E Q9.
Spent an age on this question and im still confused
The question is:
A golfball is driven from the tee with speed 30√2 meters per second at an angle α to the horizontal.
i) Show that during its flight the horizontal and vertical displacments x and y of the ball from the tee satisfy the equation:
y = x tanα - (x^2/360)(1+ tan^2 α
DONE
ii) The golf ball just clears a tree 5m high which is 150m horizontally from the tee. Find the two possible values of tanα.
DONE - ANSWERS 1.8 AND 0.6
iii) Use the discriminant of the quadratic equation in tanα to find the greatest distance by which the golf ball can clear the tree and find the value of tanα in this case.
Confused - Ok so i can understand that b^2 - 4AC must = 0 for the highest point. But how do you answer it? Its most probs very easy but my excuse is im tired lol.
Many thanks!
BTW The question is in MEI Structured Mathmatics, M1 Edition 3. Excersise 6E Q9.
are you sure you got the question right?
if you take it literally, the distance is given by maximizing (y-5)^2 +(x-150)^2 which is a bit over the top.
if it's understood that the ball passes at the top of the tree then the greatest distance might mean the greates horizontal distance -having just cleared the tree which will be one of your tan alphas.
if you take it literally, the distance is given by maximizing (y-5)^2 +(x-150)^2 which is a bit over the top.
if it's understood that the ball passes at the top of the tree then the greatest distance might mean the greates horizontal distance -having just cleared the tree which will be one of your tan alphas.
Yup, copied question straight out of the book word for word.
How did you get to (y-5)^2 +(x-150)^2 ?? Also its not greatest horizontal distance, as it says you will be calculating a new tanα and i've already calculated the greatest horizontal distance where it just clears the 5m in part ii). Im 100% positive they're talking about vertical height over the tree *i think* lol.
How did you get to (y-5)^2 +(x-150)^2 ?? Also its not greatest horizontal distance, as it says you will be calculating a new tanα and i've already calculated the greatest horizontal distance where it just clears the 5m in part ii). Im 100% positive they're talking about vertical height over the tree *i think* lol.
Solved it 
y = x tanα - (x^2/360)(1+ tan^2 α Rearranged =
0 = x tanα - (x^2/360) - (x^2/360 x tan^2 α - y
plug in x = 150 (point where it passes over tree)
0 = 150 tanα - 62.5 - 62.5 tan^2 α - y
Using the discriminant of the quadratic equation (b^2 - 4AC), we know this value must = 0 to find the highest value as it will mean there will be a double root at this point. A = -62.5, B = 150, C = -62.5 - y
So:
150^2 - 4 x -62.5 x (-62.5-y) = 0
Therefore y = 27.5 from ground and we know the tree is 5 metre tall so it clears by 22.5m. Now we know y value and x value, plug into original equation to find tanα = 1.2.
Checked answer in back of the book and its correct. Phew! Quite simple now its done but arent all Mechanics questions easy in hindsight.
y = x tanα - (x^2/360)(1+ tan^2 α
Rearranged =0 = x tanα - (x^2/360) - (x^2/360 x tan^2 α
- yplug in x = 150 (point where it passes over tree)
0 = 150 tanα - 62.5 - 62.5 tan^2 α - y
Using the discriminant of the quadratic equation (b^2 - 4AC), we know this value must = 0 to find the highest value as it will mean there will be a double root at this point. A = -62.5, B = 150, C = -62.5 - y
So:
150^2 - 4 x -62.5 x (-62.5-y) = 0
Therefore y = 27.5 from ground and we know the tree is 5 metre tall so it clears by 22.5m. Now we know y value and x value, plug into original equation to find tanα = 1.2.
Checked answer in back of the book and its correct. Phew! Quite simple now its done but arent all Mechanics questions easy in hindsight.
Thank you, looking at this question 15 years later, your working has been most helpful!
Original post by 789789
Solved it
y = x tanα - (x^2/360)(1+ tan^2 α Rearranged =
0 = x tanα - (x^2/360) - (x^2/360 x tan^2 α - y
plug in x = 150 (point where it passes over tree)
0 = 150 tanα - 62.5 - 62.5 tan^2 α - y
Using the discriminant of the quadratic equation (b^2 - 4AC), we know this value must = 0 to find the highest value as it will mean there will be a double root at this point. A = -62.5, B = 150, C = -62.5 - y
So:
150^2 - 4 x -62.5 x (-62.5-y) = 0
Therefore y = 27.5 from ground and we know the tree is 5 metre tall so it clears by 22.5m. Now we know y value and x value, plug into original equation to find tanα = 1.2.
Checked answer in back of the book and its correct. Phew! Quite simple now its done but arent all Mechanics questions easy in hindsight.
y = x tanα - (x^2/360)(1+ tan^2 α
Rearranged =0 = x tanα - (x^2/360) - (x^2/360 x tan^2 α
- yplug in x = 150 (point where it passes over tree)
0 = 150 tanα - 62.5 - 62.5 tan^2 α - y
Using the discriminant of the quadratic equation (b^2 - 4AC), we know this value must = 0 to find the highest value as it will mean there will be a double root at this point. A = -62.5, B = 150, C = -62.5 - y
So:
150^2 - 4 x -62.5 x (-62.5-y) = 0
Therefore y = 27.5 from ground and we know the tree is 5 metre tall so it clears by 22.5m. Now we know y value and x value, plug into original equation to find tanα = 1.2.
Checked answer in back of the book and its correct. Phew! Quite simple now its done but arent all Mechanics questions easy in hindsight.
Thank you, looking at this question 15 years later, your working has been most helpful!
Original post by 789789
Solved it
y = x tanα - (x^2/360)(1+ tan^2 α) Rearranged =
0 = x tanα - (x^2/360) - (x^2/360 x tan^2 α) - y
plug in x = 150 (point where it passes over tree)
0 = 150 tanα - 62.5 - 62.5 tan^2 α - y
Using the discriminant of the quadratic equation (b^2 - 4AC), we know this value must = 0 to find the highest value as it will mean there will be a double root at this point. A = -62.5, B = 150, C = -62.5 - y
So:
150^2 - 4 x -62.5 x (-62.5-y) = 0
Therefore y = 27.5 from ground and we know the tree is 5 metre tall so it clears by 22.5m. Now we know y value and x value, plug into original equation to find tanα = 1.2.
Checked answer in back of the book and its correct. Phew! Quite simple now its done but arent all Mechanics questions easy in hindsight.
y = x tanα - (x^2/360)(1+ tan^2 α) Rearranged =
0 = x tanα - (x^2/360) - (x^2/360 x tan^2 α) - y
plug in x = 150 (point where it passes over tree)
0 = 150 tanα - 62.5 - 62.5 tan^2 α - y
Using the discriminant of the quadratic equation (b^2 - 4AC), we know this value must = 0 to find the highest value as it will mean there will be a double root at this point. A = -62.5, B = 150, C = -62.5 - y
So:
150^2 - 4 x -62.5 x (-62.5-y) = 0
Therefore y = 27.5 from ground and we know the tree is 5 metre tall so it clears by 22.5m. Now we know y value and x value, plug into original equation to find tanα = 1.2.
Checked answer in back of the book and its correct. Phew! Quite simple now its done but arent all Mechanics questions easy in hindsight.
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can someone please explain what principle domain is and why the answer is a not c?Maths
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