Projectile problem (1 Viewer)

[mazza_728]

Hey guys,
This is really testing me .. can u help?
A golfer hits a ball to a green which is 15m above the elvel of projection. the angle = 30 degrees to the horizontal. the initial velocity of the ball is 36ms^-1. What was the time of the flight?
Thanks xoxo

 

[ND]

Solve s=ut+1/2*at^2 for t where s=15, u=36sin30, a=-9.8.

note: angle is in degrees (obviously).

 

[LadyMoon]

if it is above then shouldnt it be negative 15?
or it is just me? might have lost all my mathematical brain after HSC

 

[ND]

Originally posted by LadyMoon
if it is above then shouldnt it be negative 15?
or it is just me?

I think it's just you.

 

[LadyMoon]

Originally posted by ND
I think it's just you.

Oh i misread the question.
The ball is being hit to "15m above the level of projection"
Not "hit from 15m above"

Ooops

 

[freaking_out]

i'd like to say, that projectile motion in the physics course becomes a breeze when u finish the projectile motion in the 3u/4u maths course.

 

[ND]

Originally posted by freaking_out
i'd like to say, that projectile motion in the physics course becomes a breeze when u finish the projectile motion in the 3u/4u maths course.

Well it wasn't exactly hard before.

 

[mazza_728]

obviously it is hard.. cause by doing that i get the wrong answer i get like 1.15 when its suppose to be 2.4 -- can someone else try?

 

[Arell]

this is a little off topic, but its on projectiles
we have a prac coming up, any idea what it could be on, formulas, and any tips
thx

 

[Arell]

oh, well if you took it as a two part question it'd be easeir, think about it, do how long it takes to get to the topmax height, and find the max height, subtract the fifteen, and find how long it takes to fall in the second section, genius

 

[ND]

Originally posted by mazza_728
obviously it is hard.. cause by doing that i get the wrong answer i get like 1.15 when its suppose to be 2.4 -- can someone else try?

I couldn't be bothered actually doing the question (i've finished my hsc, i'm allowed to be lazy ), but it looks to me (by looking at the equation) like both 1.15 and 2.4 are roots, and the larger root is obviously the one you're after.

 

[ND]

Originally posted by Arell
this is a little off topic, but its on projectiles
we have a prac coming up, any idea what it could be on, formulas, and any tips
thx

Just know the usual formulas (i.e. s=ut+1/2at^2 and its derivative and variations etc.). It could be on observering the path of a projectile (with a data logger or camera or something).

 

[mazza_728]

AHHH! kill me now! ok i did that and i got that time in the first half is 1.84. and then coming down --- would initial velocity equal zero?? cause at maxiumum height there is not velocity?? does anyone understand me? so then in the 2nd half i got that t=2.66 which all up would be like 4. something??? incorrect :'(

 

[ND]

Just solve the quadratic in my first post... it'll give you the correct answer.

 

[freaking_out]

Originally posted by ND
Well it wasn't exactly hard before.

i didn't say that it was hard bfore.

yeah, note the trick of using the second equation of motion to form a quadratic- it serves as a shortcut alot of the time.

 

[babii > girl]

can someone please help me with these projectile motion question??? im sooo lost... and my assignment is due on monday.... *stressin*

1) How does initial velocity influence range and maximum height?
2) how does angle of trajectory influence range and max. height?

 

[Constip8edSkunk]

let u = initial velocity v=final velocity a=acceleration s = displacement @=angle of projection

v^2 = u^2 +2as
s = [v^2-u^2]/2a ------ *
now consider only the vertical vectors.

initially u_vert. = u*sin@ {by dividing the initial vector into hor. and vert. vectors and using trigonometry}

at maximum height, v_vert. = 0

s then represent the maximum height, H

so from * , H = [ 0 - usin@ ]/2a

but vertically the acceleration is g (assuming no air resistance)

.'. H = usin@/2g

now g is assumed 2 b constant in projectile motion, leaving only u and @ as variable, both of which are directly proportional to H, the maximum height.

so assuming each value is varied at a time, the maximum height increases linearly with initial velocity, and also linearly with the the angle of projection (0<@<pi/2)

 

[ajyt]

mazza
when u finish mechanics in 4 unit maths, motion and forces will make a lot more sense.

the physics motion equations automate the integration process for paths beginning at the origin.

 

[mazza_728]

thankyou .. im hoping it all becomes clearer

 

[zeropoint]

Originally posted by babii > girl
can someone please help me with these projectile motion question??? im sooo lost... and my assignment is due on monday.... *stressin*

1) How does initial velocity influence range and maximum height?
2) how does angle of trajectory influence range and max. height?

The initial velocity u of a projectile can be decomposed into a component u_x in the horizontal direction and a component u_y in the vertical direction. These are related by the equations

u_x = u cos@
u_y = u sin@
u^2 = (u_x)^2 + (u_y)^2

The horizontal component of velocity determines the horizontal range x while the vertical component of velocity determines the maximum height y attained. Assuming the object is projected from the origin,

x = ut cos@
y = ut sin@ − (1/2) g t^2

It can be quickly observed that y is maximised when @ = 90 , that is, when it is projected straight up. To obtain the maximum range, we first note that

v_y = u sin@ − gt

where v_y is the instantaneous vertical component of velocity at any time t. Letting v_y = −u sin@ (the velocity when the projectile strikes the ground) we find that

t = (2u sin@)/g

which is maximised when @ = 90. Therefore x is maximised when @ = 45.