Projectile Motion Question (1 Viewer)

[joshisthebest]

here...


Don't really know what to do... thanks for any help.

 

[香港!]

taking gravity, g=-10

Consider the horizontal components:
x''=0, x'=60cos45=60\sqrt2, x=60t\sqrt2
Consider the vertical components:
y''=-10, y'=-10t+60\sqrt2, y=-5t²+60t\sqrt2+50

set y=0, 5t²-60t\sqrt2-50=0
t=[60\sqrt2 +-sqrt(60²\2-4(5)(-50))]\10
=[60\sqrt2+-sqrt 2800]\10
t=9.534 by calculator, ignoring the negative value.
So it takes t=9.534 seconds to reach the sea
at this time, the particle would have travelled a distance given by
x=60(9.534)\sqrt2=404.49 m by calculator
that's the required "range"

 

[SlaminSammy]

this needs to be done in 5 steps

first consider vetrically

time to reach max height ----- take up to be negative

v = u + at
0 = 60sin45 -9.8 x t
t = 4.33 s

max height

v^2 = U^2 + 2as
0 = (60sin45)^2 - 19.6 x s
s = 91.84 m

hence vertical displacement = 91.84 + 50 = 141.84m

velocity before hits ground

v^2 = u^2 + 2as
= 0 - 19.6 x 141.84
v = -52.73 m/s

time from max height to ground

v = u +at
-52.37 = 0 + -9.8 t
t = 5.38

hence total time = 5.38 + 4.33 = 9.71

consider horizontally

s = ut + 0.5at^2
s = 60cos45 x 9.71
s = 411.96m

therefore
range = 411.96m

of course you can use s = ut + .5at^2 to figure out time from max height to ground and take out a step but this is out of my league. :uhhuh:

 

[exa_boi87]

I used a similar approach to both here .. however only in 2 steps ..

taking 9.8 as acceleration due to gravity ..

s[y] = u[y]t - .5at^2
-50 = 60sin45t - 4.9t^2
-50 = 60t/(sqrt 2) - 4.9t^2
4.9t^2 - 60t/(sqrt 2) - 50 = 0
Quadratic Forumlua (too much to type lol) but you know the drill -b +- (sqrt)(b^2-4ac)/(2a)
You get two values out, one is negative and the other is -> 9.709... seconds

Then, change in horizontal range = u[x]t = 60cos45 x 9.709 ... = 411.93m

Now that text book says 411 as the answer (as I use it as well) however we dont know if theyve taken g to be 9.8 or 10, and this textbook has a nice habit of rounding off answers (take a look at the previous questions answer)

edit: Just tried g = -10 and 404.499 ... the same answer that was calculated previously by another member

 

[joshisthebest]

Thanks for the help

 

[Haku]

I used a similar approach to both here .. however only in 2 steps ..

taking 9.8 as acceleration due to gravity ..

s[y] = u[y]t - .5at^2
-50 = 60sin45t - 4.9t^2
-50 = 60t/(sqrt 2) - 4.9t^2
4.9t^2 - 60t/(sqrt 2) - 50 = 0
Quadratic Forumlua (too much to type lol) but you know the drill -b +- (sqrt)(b^2-4ac)/(2a)
You get two values out, one is negative and the other is -> 9.709... seconds

Then, change in horizontal range = u[x]t = 60cos45 x 9.709 ... = 411.93m

Now that text book says 411 as the answer (as I use it as well) however we dont know if theyve taken g to be 9.8 or 10, and this textbook has a nice habit of rounding off answers (take a look at the previous questions answer)

edit: Just tried g = -10 and 404.499 ... the same answer that was calculated previously by another member

um? can we actually use the quadratic formula from math in physics? as i my self do not really see a reason not to, but just making sure.

 

[helper]

You can use any appropriate formula, including the maths methods

 

[Haku]

thanks helper.

helper, any final tips for use physics hsc people?

 

[helper]

I'll try and write something up over the weekend.

 

[Haku]

thanks. maybe post it in a new thread, the mods could probably make it a sticky.