- #51

- 322

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter imy786
- Start date

- #51

- 322

- 0

- #52

cristo

Staff Emeritus

Science Advisor

- 8,107

- 73

Why? You are given sufficient information to answer each part of the question, so there is no need to enter random figures!

- #53

hage567

Homework Helper

- 1,509

- 2

Why would you do that? You will have to solve all of it eventually. If you do it in the right order it makes everything easier. It will also be less confusing if you don't jump around from one thing to another.

- #54

- 322

- 0

component form.

--------------------------------------------------------------------------

2 equations- horizontal and verical motion-

horizontal speed remains constant as there is no acceleration

horizontal speed is= 20cos 11.53= 19.60m/s

vertical motion to calculate vertical speed:

v=

u= 20sin11.52= 4m/s

a= -10

t=2 (time from part c calculated solving quadratc equaiton)

v=u+at

= 4- (20)= -16 m/s = vertical speed

???/

is this vertical speed correct ?

- #55

- 322

- 0

need help here............

- #56

- 322

- 0

please someone help me do this quesiton URGENT.....

- #57

hage567

Homework Helper

- 1,509

- 2

component form.

--------------------------------------------------------------------------

2 equations- horizontal and verical motion-

horizontal speed remains constant as there is no acceleration

horizontal speed is= 20cos 11.53= 19.60m/s

vertical motion to calculate vertical speed:

v=

u= 20sin11.52= 4m/s

a= -10

t=2 (time from part c calculated solving quadratc equaiton)

v=u+at

= 4- (20)= -16 m/s = vertical speed

???/

is this vertical speed correct ?

Yeah, that seems OK for the vertical velocity. So put that together with the horizontal velocity to get the total velocity. Use the Pythagorean theorem. Please look it up if you don't know how to do it, and give it a try.

please someone help me do this quesiton URGENT.....

Then please work everything out to your final answers for your next post so it can be checked all at once. This question seems to have been considered "urgent" for quite some time.

- #58

- 322

- 0

component form.

--------------------------------------------------

total velocity= adding both vector velocity

horizontal=19.60 m/s

vertical speed= -16

inverse tan (-16/19.6) = angle= -39.22

total velocity= horizontal +vertical = 19.6- 16 =3.6 m/s

magnitude of velocity is 3.6 m/s at angle -39.2 degrees direction.

is this correct?

angle being a minus ???should that be ok....????

- #59

hage567

Homework Helper

- 1,509

- 2

component form.

--------------------------------------------------

total velocity= adding both vector velocity

horizontal=19.60 m/s

vertical speed= -16

inverse tan (-16/19.6) = angle= -39.22

total velocity= horizontal +vertical = 19.6- 16 =3.6 m/s

magnitude of velocity is 3.6 m/s at angle -39.2 degrees direction.

is this correct?

angle being a minus ???should that be ok....????

A negative angle is fine, think about which direction the velocity is with respect to the horizontal at the end of the jump.

NO the total velocity is not 3.6 m/s. I told you to use the Pythagorean theorem, do you not know what that is? How do you find the hypontenuse of a right angle triangle? That is really what you are doing here.

- #60

- 322

- 0

a2+b2= c2...

hage- iv done a leve maths and physics ...so i aint a total beginer...

sqaure root of (19.6^2 + 16^2) = 25m/s is the total velocity at -39.2 degrees

with respect to the horizontal

- #61

hage567

Homework Helper

- 1,509

- 2

a2+b2= c2...

hage- iv done a leve maths and physics ...so i aint a total beginer...

sqaure root of (19.6^2 + 16^2) = 25m/s is the total velocity at -39.2 degrees

with respect to the horizontal

Well when I say use Pythagorus and you don't, what else am I to assume!?

That answer looks reasonable to me.

- #62

- 322

- 0

you have been so helpful...........

Share: