Projectile Motion killing us all (1 Viewer)

[nick1048]

projectile motion

all questions with 1 dot are do-able but I can't seem to get them out please help

double dot means these are hard... thanks

btw any help would be really really appreciated *<3 to all!*

 

[acmilan]

http://www.boredofstudies.org/community/showthread.php?t=35496&highlight=projectile

Answers one of those you are having trouble with. Good old CM_Tutor

 

[Pace_T]

For 12, See attachment. However for the 2nd part, I did the same method but I got h(b^2-c)/[bc(b-c)]

Since 15 is already done

Here is 16.
A) y = -gt^2 +vtsin@
when t= 2, y = 0

2vsin@ = 20
vsin@ = 10 _______1
now, x = vtcos@
when t = 2, x = 50
25 = vcos@ _______2

1 over 2

tan@ = 10/25
@ = 21.8

.'. from ____2, v = 25/cos21.8 = 26.93m/s

B)for max height, t = 1 (since the launch and landing is the same level, the height is the middle since it is a parabola. so therefore max height occurs half of time of flight)

.when t = -1,
y = -5 + 26.93sin(21.8)
=5m

 

[Pace_T]

18 and 19 in attachment

 

[Pace_T]

heres 20

umm sorry for the crappy fotos and handwriting, i cbf typing it up and it would just seem even more confusing ,_,

its better than nothing i guess

 

[shafqat]

For 12, sub y = 7 into the equation of flight, then use sums and products of roots in the quadratic in x, as the two roots will be 7 and 14. This method should give you the general expression too Pace_T

 

[nick1048]

thanks pace, can't give you any more rep :\ lol I got the second part of 12... I'm still stuck on 17, 22 and 24 but the last two arent scanned up. That'll do I'm sick of maths atm lol, thanks to you all very very much.

 

[Pace_T]

No worries dude, my pleasure
As for second part of twelve, how did you work that one out? Same method or did you use Shafqat's method?
I'll have a look at those q's for you.
Cheers,
Pace T.

 

[Pace_T]

17) http://www.boredofstudies.org/community/showthread.php?t=35496 it's near the last post.
Although, I don't how he done it lol.. I am missing something, I might have to ask the teacher :O

22)
28 = vcosa_____1
v^2sin^2a/2g = 4.9
v^2sin^2a = 96.04______2
from 1, v = 28/cosa
v^2 = 28^2/cos^2a _____3

3--> 2
28^2/cos^2a * sin^2a = 96.04
tana = 7/20
use the triangle,
cosa = 20/sqrt449
sina = 7/sqrt449
v= 28*sqrt449/20
= 7sqrt449/5

now, when y = 1.3,
1.3 = -4.9t^2 + vtsina
= -4.9t^2 + 7sqrt449/5 * 7/sqrt499 * t
= -4.9t^2 + 49t/5
t = [49 +/- sqrt(96.04 - 25.48)]/9.8
t = (9.8 +/- 8.4)/9.8
t = 1.857 or t = 0.1428 (pick the later time as this is when the ball is returning to the x axis)
when t = 1.857, x = 1.85* 20/sqrt449 * 7sqrt499/5
x = 52,
distance to run is 65-42 = 13m in 1.85 seconds
v = d/t
= 13/1.85
= 7m/s

 

[nick1048]

Thanks again man your like the maths ace lol I don't know how you do it hahahah
attached is question 12b starts from the diagram, please excuse the messy handwriting, thanks

 

[Jumbo Cactuar]

Hey I'll explain Q12 a bit better

blue is supposed to be the parabolic path.

The distance between the pole and the bird is 10 times the time (10 t)
The distance between the person and the bird is V_h (the horizontal velocity of the stone) times the flight time ( V_h.t )

The distance between the pole and the maxima of the parabola is half of 10 times the time. (5 t)
The distance between the person and the maxima of the parabola is V_h (the horizontal velocity of the stone) times the flight time - the distance between the bird and the maxima of the parabola. (V_h.t - 5t)

We know that the basic parabola has the function y=nx², and so y/x² is a constant for the parabola. We know the points (-h,5t) and (-2h,V_h.t - 5t) are on the parabola. So;

(5t)²/h = (V_h.t - 5t)² / 2 h
2(5)²= (V_h -5)²
50 = 25 - 10 V_h + V_h²
V_h² - 10 V_h -25 = 0

V_h = (10 +/- sqrt(100+100))/2
= 5 (1 +/- sqrt(2))
but V_h is larger than 10 so;
V_h = 5 (1 + sqrt(2))