Projectile question? From Fitzpatrick 3U (2 Viewers)

Vidhya · Oct 4, 2011

Question 24 from excercise 25 (d)

Find the speed and direction of a particle which, when projected from a point 15m above the horizontal ground, just cleared the top of a wall 26.25m high and 30m away.

I've tried a few things but the numbers are getting really messy and it's not working out!

I'm also wondering if this is a generic question in the HSC, or if they are a tad easier?

Thank you!

Answer: 25 m/s, 36 degrees 52'

apollo1 · Oct 4, 2011

Vidhya said:
Question 24 from excercise 25 (d)

Find the speed and direction of a particle which, when projected from a point 15m above the horizontal ground, just cleared the top of a wall 26.25m high and 30m away.

I've tried a few things but the numbers are getting really messy and it's not working out!

I'm also wondering if this is a generic question in the HSC, or if they are a tad easier?

Thank you!

Answer: 25 m/s, 36 degrees 52'

lol this question is a million times easier than wat u will get in the hsc.

bleakarcher · Oct 4, 2011

is there enough information to solve for the two variables?

Vidhya · Oct 4, 2011

bleakarcher said:
is there enough information to solve for the two variables?

Yes :S I made simultaneous equations using (0,15) and (30,26.25). Is that the right way to go?

bleakarcher · Oct 4, 2011

well if ur using those points that implies that u used the ground 15m below the point of projection as the origin. the equations of motion, as a result would have been:
y=-5t^2+Vtsin(A)+15, x=Vtcos(A)
Now, y=-5[x/(Vcos(A))]^2+Vtsin(A)+15=xtan(A)-5x^2/[V^2cos^2(A)]+15
if u use the point (0,15):
0+0+15=15
15=15, it doesnt get u anywhere lol. as if u didnt realise we have already used the point in order to find the equation for the vertical component of motion as an initial condition.
however, the second point u can use to relate V and A in one equation. however to solve for the two variables u need 2 equations. as there is no additional data, u can not solve for V nor can u for A

Vidhya · Oct 4, 2011

Yeah I realised 0,15 wouldn't work

. So we've only got one set of points for the equation. There's definitely no way of solving this then?

bleakarcher · Oct 4, 2011

yeh lol, u cant solve for the question. maybe they forgot to give the point in time when which the particle just cleared the wall or something like that.

Vidhya · Oct 4, 2011

bleakarcher said:
yeh lol, u cant solve for the question. maybe they forgot to give the point in time when which the particle just cleared the wall or something like that.

Phew, okay so I'm not all that hopeless at maths haha. Thanks

taeyang · Oct 4, 2011

Wow that was a pretty difficult question! nah man they don't ask harder than that in the HSC. ^.^
$$Max height when$\, \dot{y}=0 \\ \therefore -gt+vsin\alpha = 0\\ \therefore t = \frac{vsin\alpha}{g}\\ $Now, Max height$\\ = \frac{v^2sin^2\alpha}{2g}+15\\ \therefore \frac{v^2sin^2\alpha}{2g}+15 = 26\tfrac{1}{4}\\ \therefore \frac{v^2sin^2\alpha}{2g}=11\tfrac{1}{4}\\ \therefore v^2sin^2\alpha = \frac{45}{4}.2.10\\\\ v^2sin^2\alpha = 225\\ vsin\alpha = 15\\\\ $Max Height occurs when$\, t = \frac{vsin\alpha}{g}\\ = \frac{15}{g}\\ = \frac{15}{10}\\ = \frac{3}{2}\\ $At$\; t = \frac{3}{2}, x = 30\\ \therefore vtcos\alpha = 30\\ v.\frac{3}{2}.cos\alpha = 30\\ vcos\alpha = 20\\\\ \therefore \frac{vsin\alpha}{vcos\alpha}=\frac{15}{20}\\ \therefore tan\alpha = \frac{3}{4}\\ \therefore alpha=36^{\circ}52{}'\\ \therefore vsin\alpha = 15\\ \therefore v.\frac{3}{5}=15\\ \therefore v = 25m/s$

Vidhya · Oct 5, 2011

Ohhhhhhhhhh wow, genius! Thank youu I appreciate that

)

Kyrix · Oct 5, 2011

taeyang said:
Wow that was a pretty difficult question! nah man they don't ask harder than that in the HSC. ^.^
$$Max height when$\, \dot{y}=0 \\ \therefore -gt+vsin\alpha = 0\\ \therefore t = \frac{vsin\alpha}{g}\\ $Now, Max height$\\ = \frac{v^2sin^2\alpha}{2g}+15\\ \therefore \frac{v^2sin^2\alpha}{2g}+15 = 26\tfrac{1}{4}\\ \therefore \frac{v^2sin^2\alpha}{2g}=11\tfrac{1}{4}\\ \therefore v^2sin^2\alpha = \frac{45}{4}.2.10\\\\ v^2sin^2\alpha = 225\\ vsin\alpha = 15\\\\ $Max Height occurs when$\, t = \frac{vsin\alpha}{g}\\ = \frac{15}{g}\\ = \frac{15}{10}\\ = \frac{3}{2}\\ $At$\; t = \frac{3}{2}, x = 30\\ \therefore vtcos\alpha = 30\\ v.\frac{3}{2}.cos\alpha = 30\\ vcos\alpha = 20\\\\ \therefore \frac{vsin\alpha}{vcos\alpha}=\frac{15}{20}\\ \therefore tan\alpha = \frac{3}{4}\\ \therefore alpha=36^{\circ}52{}'\\ \therefore vsin\alpha = 15\\ \therefore v.\frac{3}{5}=15\\ \therefore v = 25m/s$

Isnt this assuming that the maximum height is at the fence?

but if you assume its max height you get the answer so... Imo, this is a bad question.

The projectile in my experience are generally harder than this question (if they told you it was max height)

bleakarcher · Oct 5, 2011

yeh i dont think u can assume max height occurs at the point where it clears the wall. the question doesnt say

Vidhya · Oct 5, 2011

bleakarcher said:
yeh i dont think u can assume max height occurs at the point where it clears the wall. the question doesnt say

Most questions in this textbook are worded badly, so I'm gonna assume that they meant it was the max height

Vidhya · Oct 5, 2011

Kyrix said:
The projectile in my experience are generally harder than this question (if they told you it was max height)

So is this considered an easy-ish question? How hard can they get? I've done a few past questions and I could do all of them, which says something because I am le miserable at maths.

OldMathsGuy · Oct 5, 2011

bleakarcher said:
yeh i dont think u can assume max height occurs at the point where it clears the wall. the question doesnt say

The question is confusing unless you think about it as:
Imagine you fire the ball at a set velocity and angle so it just clears the fence but not when it is at the apex of its path. Using this same velocity, you could choose an alternative angle so that it easily clears the fence. And so when it says "just" clears the fence, it ambiguously means that where it just clears the fence is the only way how it could just clear the fence. When the ball clears the fence at the apex of its path, any change in angle or decrease in velocity will mean it does not clear the fence and so this is the condition that restricts the answer.

Very ambiguous though - could have been much more clearly presented. Generally though, projectile questions on the HSC papers exceed this in difficulty if clearly presented (which they typically are). I thought the recent one that used applied knowledge of roots was a neat question.

Best Regards
OldMathsGuy

K4M1N3 · Oct 5, 2011

bleakarcher said:
yeh i dont think u can assume max height occurs at the point where it clears the wall. the question doesnt say

I don't believe that taeyang used max height as the point where the projectile clears the wall. The max height is just used to help define variables of the parabola to then determine V and /alpha using the information given as the projectile clears the wall.

Projectile question? From Fitzpatrick 3U (2 Viewers)

Vidhya

Member

apollo1

Banned

bleakarcher

Active Member

Vidhya

Member

bleakarcher

Active Member

Vidhya

Member

bleakarcher

Active Member

Vidhya

Member

taeyang

Member

Vidhya

Member

Kyrix

New Member

bleakarcher

Active Member

Vidhya

Member

Vidhya

Member

OldMathsGuy

Premium Member

K4M1N3

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)