projectile motion (1 Viewer)

[marsenal]

A ball is projected so that its horizontal range is 45 metres and it passes through a point 22.5 metres horizontally from and 11.25 metres vertically above the point of projection. Find the angle and speed of projection.

 

[kini mini]

Once you have the range you can just prove and use the range formula. This can be helpful where the second piece of info isn't as good.

But in this case, we say by symmetry that since we know the position of the ball midway through its flight, we know where V(y) = 0, etc etc

Start with the usual x = V cos @, y = V sin @. Then integrate using the info to determine the constant terms.

 

[Affinity]

Approach:

Derive the location of the particle in parametric form, with time as the parameter, i.e x and y in terms of t.

solve these 2 equations for the range and maximum height of the particle, noting that at half the range, (22.5) the particle is at it's maximum altitude

=======================================
let a_y, a_x, v_y, v_x be the acceleration and velocity functions of time of the particle

a_y = -g (gravity)
a_x = 0
v_y(0) = v*sin(q) where q is the angle of projection
v_x(0) = v*cos(q) and v is the speed of projecton
y(0)=0
x(0)=0

now

/
| a_x dt = v_x
/

C = v_x

using initial values, C= v*cos(q)

v_x = v*cos(q)

/
| v_x dt = x
/

vt*cos(q) + C = x

again using initial values, C = 0

vt*cos(q) = x


/
| a_y dt = v_y
/

-gt + C = v_y

using initial values, C= v*sin(q)

v_y = v*sin(q) - gt

/
| v_y dt = y
/

vt*sin(q) - (g/2)*t^2 + C = y

using initial values, C = 0

y = vt*sin(q) - (g/2)*t^2



x = vt*cos(q)
y = vt*sin(q) - (g/2)*t^2

y = t(v*sin(q) - (g/2)t)
y = 0 when t = 0 or (2v/g)*sin(q)

therefore range = v*(2v/g)*sin(q)*cos(q) = (2v^2/g)sin(q)cos(q)

(2v^2/g)sin(q)cos(q) = 45 (*)

maximum height = max value of y
y is a quadratic in t, therefore must take it's max value when
t = -[v*sin(q)]/[2*(-g/2)]
t = (v/g)sin(q)

max height = v*(v/g)sin(q)*sin(q) -(g/2)[(v/g)sin(q)]^2
=(v^2/g)[sin(q)]^2 - [(v^2)/(2g)][sin(q)]^2
=[(v^2)/(2g)][sin(q)]^2


[(v^2)/(2g)][sin(q)]^2=11.25 (**)

now divide * by **

4cot(q)=4
cot(q)=1
q=pi/4 (45 degrees)

from *

(2v^2/g)/2= 45

v=sqrt(45g)= 3*sqrt(5g)

 

[jayz]

Mr Chan, i really think u should spend more time studying eng than polishing up your mathematical skills! (no offence)

 

[wogboy]

Mr Chan, i really think u should spend more time studying eng than polishing up your mathematical skills! (no offence)

Who's Mr Chan?

 

[Affinity]

Jayz ... adhere to internet conventions, I am identified with my screen name here.

 

[jayz]

I don't think i gave away anyone's identity!

Oops, i just realised (somebody) gave it way themeselve, they must be in 'deep' trouble! (this is suppose to be ironic)

Any way even if your name is given on the net, no one would realise if it's really true, until they actually see your identifications!

So wat's this busisness about net conventions?

 

[Affinity]

going off topic here.

Jayz: I don't mind, but some of your other friends do mind if you post their real name on forums.

and I am too young to be a Mr.. don't mislead people.