Projectile motion (clueles)

  • #1

Homework Statement


A jet fighter of mass 3t climbs from ground level to a height of 10,000m
in one minute. At that time it has a horizontal component of velocity
of 241m/s and a vertical component of velocity of139m/s. find
(A) the direction in which it is travelling
(B) the speed
(C) its kinetic enregy
(D) its increase in gravitational potential energy
(E) the power delivered by its engines during the climb


2. Relevant
I've only been doing this for 3 weeks now, and i get some of it
but dont get most.
anyhelp would be nice..., i get that theres alot.
but im sure its basic stuff, im not looking for answers
just help.


The Attempt at a Solution



My attempt

a) tan = Vx/Vy
=139/241
=0.5767
=30 degrees

just a tidbit, i know that tan(29.96 =0.5767 But how do i get 0.5767 to = 29.96?)

b) V²=Vy²+Vx²
=241²+139²
=77402
=278.2121 m/s

The answers said it = 27.8m/s, why?



the next three are where i
have even less of an idea.

c) KE=½mv² right
=0.5x3000x278.2121
=417318.15j, but thats too high,right?

the answer book said it =1.16x108j , which
freaked me out because what is with the
x? where did that come from!


d) is it to do with PE=mgh? cause i did that, but no go

e) righteo.

(sorry for any spelling errors)
 
Last edited:

Answers and Replies

  • #2
cristo
Staff Emeritus
Science Advisor
8,107
73

Homework Statement


A jet fighter of mass 3t climbs from ground level to a height of 10,000m
in one minute. At that time it has a horizontal component of velocity
of 241m/s and a vertical component of velocity of139m/s. find
(A) the direction in which it is travelling
(B) the speed
(C) its kinetic enregy
(D) its increase in gravitational potential energy
(E) the power delivered by its engines during the climb


2. Relevant
I've only been doing this for 3 weeks now, and i get some of it
but dont get most.
anyhelp would be nice..., i get that theres alot.
but im sure its basic stuff, im not looking for answers
just help.


The Attempt at a Solution



My attempt

a) tan = Vx/Vy
=139/241
=0.5767 (just a tidbit, i know that tan(29.96 =0.5767
=30 degrees But how do i get 0.5767 to = 29.96?)
You have [itex]\tan\theta=0.5967 \Rightarrow \theta=\tan^{-1}(0.5967)[/itex]

b) tan = V²=Vy²+Vx² The answers said
=241²+139² it = 27.8m/s, why?
=77402
=278.2121 m/s
I don't know you've got the tan there. The correct equation is V2=Vy2+Vx2. Hence, V2=77402, and thus V=√(77402)

Incidentally, is the bit I've highlighted in blue a typo, and should read 278?

the next three are where i
have even less of an idea.

c) KE=½mv² right
=0.5x3000x278.2121 so to me its goto be
=417318.15j, but thats too high?
Why do you think that? It is incorrect though, since you used the value for v instead of the value for v2

the answer book said it =1.16x108j , which
freaked me out because what is with the
x? where did that come from!
This means T=1.16x108=116000000

d)is it to do with PE=mgh?
Yes.

e) righteo.
??
 
  • #3
Oh, the tan was a mistake
& im a goob, i didn't get that T=1.16x108=116000000
stuff makes more sense, cheers.
 

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