Calculating velocity (1 Viewer)

[rawker]

A long jumper left the ground at an angle of 15degrees and jumped a distance of 7.0m. Considering this result, caluculate the velocity of long jumper when leaving the ground.



How would I go about answering that?

 

[gordo]

work out the height he went using triganometry (7*tan15). so u know he went that high, it took half the time to get to that height and the 2nd half of the time to come from that height back to the ground (cause vertical net force is only gravity) therefore u know that half the time in the air was 7*tan15 / 9.81 so total time in the air was twice that. Now u know how long he was in the air and how far he went (vertically and horizontally), the rest u can work out

 

[richz]

write down every info they give u. Look at the eqns and sub in.

 

[simbim21]

what's the answer?

i got 18.9m/s, but that sounds a bit wrong.

 

[Riviet]

work out the height he went using triganometry (7*tan15)

Wouldn't the angle from the starting point to the max height be less than 15 degrees due to gravity?

 

[richz]

ok, i tried it

using x=ut
7=vcos15*t
this will find the time of flight
y=0 at end pt this is when t = time of flight ie. 7/(vcos15)

so we sub this into the eqn: y=ut+.5at^2

y=0 so 0=(vsin15)*7/(vcos15) -4.9 (7/vcos15)^2

so work out v, which shud be 11.71m/s

 

[sasquatch]

Erm.. well i did this...

7 = u_xt
t = 7 / u_x

At both the starting and finishing points, r = 0

Using r = ut + (1/2)at²

0 = u_yt + (1/2)(-9.8)t²
0 = t(u_y - 4.9t)

t = 0 or u_y - 4.9t = 0
hence t = 0 or t = u_y / 4.9t

but t = 7 / u_x

therefore, 7 / u_x = u_y / 4.9t

u_xu_y = 34.3

u_x = v cos 15
u_y = v sin 15

then, v cos 15 * v sin 15 = 34.3
v² = 34.3 / (cos 15 * sin 15)
v² = 137.2

therefore v = 11.71 ms^-1 at 17^o to the horizontal




I tested it out by:

- Without basing on the information (apart the angle) given in the question:

u_y = 11.7 sin 15
t = 11.7 sin 15 / 4.9t
= 0.617 s (3. d.p)

/\x = 11.7 cos 15 * 0.617 s
= 6.984 m (3.d.p) so i think yeah its right..

 

[insert-username]

Wouldn't the angle from the starting point to the max height be less than 15 degrees due to gravity?

Projectile motion assumes no wind resistance or other pesky effects that throw calculations off. Besides, if the angle was "less than 15 degrees", what exactly would it be? It's just too hard to work with.


I_F

 

[richz]

Erm.. well i did this...

7 = u_xt
t = 7 / u_x

At both the starting and finishing points, r = 0

Using r = ut + (1/2)at²

0 = u_yt + (1/2)(-9.8)t²
0 = t(u_y - 4.9t)

t = 0 or u_y - 4.9t = 0
hence t = 0 or t = u_y / 4.9t

but t = 7 / u_x

therefore, 7 / u_x = u_y / 4.9t

u_xu_y = 34.3

u_x = v cos 15
u_y = v sin 15

then, v cos 15 * v sin 15 = 34.3
v² = 34.3 / (cos 15 * sin 15)
v² = 137.2

therefore v = 11.71 ms^-1 at 17^o to the horizontal




I tested it out by:

- Without basing on the information (apart the angle) given in the question:

u_y = 11.7 sin 15
t = 11.7 sin 15 / 4.9t
= 0.617 s (3. d.p)

/\x = 11.7 cos 15 * 0.617
= 6.984 m (3.d.p) so i think yeah its right..

nope thats rite, i think i did sum mathematical error

opps, i know where i went rong, i forgot to square 7 , how doppy of me :'(

 

[sasquatch]

man i dont know what you guys are talking bout with this angle to the top...

hehe look how long mine is compared to yours.. oh well

 

[richz]

lol, i cbf typing so much, so i just removed most of the work i did on the calc

 

[sasquatch]

i see.. i like to write down mostly everything i think (relating to physics and maths) i always draw pictures too so i can clearly see my thoughts on paper..

 

[richz]

lol, maybe thats y i make so many silly mistakes

So for future students, heres my 2cents:

- Read the q
- Draw a diagram
- Write all the info down
- work out the components
- determine which eqns to use
- sub it in
- dont skip steps