parabola question....can u solve this easy q.? :) (1 Viewer)

[lillaila]

Ive been gettin diff answs.
Can u do this so I can pick out which answ of mine is right by comparing it wif urs.

Q: Jim hits ball with v of 60km/h, striking ball from 0.9m off ground. Thru what angles should he hit ball so that it just clears a 3m high wall that's 18m away. Use 9.8 for g.

thnks

 

[evilc]

i had a crack at it and got the two possible values of theta as:
69.177 degrees, and, 27.477 degrees.

I subbed both values of theta into the cartesian equation for trajectory and got y=3, when x=18. so I'm pretty sure that my solution is correct

 

[kimmeh]

i did the same thing.. sub into the trajectory equation
i got:
68.6023 degrees and 30.85 degrees

dont u have to change it to m/s
ie 60 km/h = 16.67 m/s ?

 

[evilc]

i changed it to m/s and left it as 16 2/3 m/s, maybe that could have an effect on the answer, I'll have another go at the question, i havent done projectiles for almost a year, so i could be a bit rusty. Did you remember to take into account that the projectile was launced from 0.9m off the ground?

 

[evilc]

i did the question yet again, but didnt include the 0.9m off the ground that the projectile is launched from, doing this i got the same answers as you. So im pretty sure that my original answers are correct

 

[lillaila]

yep. thanks so much pple...that's right

 

[freaking_out]

Originally posted by lillaila
....can u solve this easy q.?

yes i can.

 

[kimmeh]

Originally posted by lillaila
yep. thanks so much pple...that's right

which one is correct ? mine or evilc's?

lol yeah i did the launching from 0.9 m above the ground...

 

[jogloran]

I got evilc's answers.

 

[kimmeh]

>.<
what did i do wrong then?
i did exactlty what he did

 

[jogloran]

Actually to tell the truth, I got the quadratic, thought "Stuff it, way too many decimals", and used a solving program.

 

[evilc]

kimmeh, did you interpret the question as the top of the wall being 3 metres above the launch point of 0.9 metres? because if you did that, you would have gotten an incorrect answer, because the wall is 3 metres off the ground, not 3.9 metres off the ground. I'll post the quadratic equation that is solved to get tan(theta)

 

[evilc]

let t be tan(theta)

5.71536t^2 - 18t + 7.81536 = 0

t = {18 +/- SQRT(324 - 4(5.71536*7.81536)} / (2*5.71536)


t = 2.62934 (5dp) --> theta = 69.177 degrees (3dp)

OR

t = 0.52007 (5dp) --> theta = 24.478 degrees (3dp)

 

[kimmeh]

Originally posted by evilc
kimmeh, did you interpret the question as the top of the wall being 3 metres above the launch point of 0.9 metres? because if you did that, you would have gotten an incorrect answer, because the wall is 3 metres off the ground, not 3.9 metres off the ground. I'll post the quadratic equation that is solved to get tan(theta)

hmm i cant remeber....wait.. ithink i put + 3.9 for the end of the y component shit.. i'm lucky this wasnt the HSC