projectile motion question (1 Viewer)

[brotha.d]

i need some help with this question

a model rocket is launched from the earth at and angle of 55 degrees from the vertical with an initial velocity of 120m/s. what is the projectiles speed 3seconds later?
a) 68.8m/s
b)98.3m/s
c)106m/s
d) 120m/s

i got the two components of motion to be
uh = 68.83m/s
uv = 98.3m/s and uh stays constant but do i need to combine these two together again and how?

thanks.

 

[Adequate]

Erm.. do u have da answer for dat? i got (D),

if da answer is (d) den i can send u my working out for it

 

[brotha.d]

answer says b

can u show me what u did?

my main question is how do u recombine the 2 components?

 

[brotha.d]

but seeing as its been shot into the air and gravity which is a negative acceleration of 9.8m/s is there, shouldnt it be different to the initial velocity?especially as they included the 3sec bit.

 

[Adequate]

er lol..... my beautiful diagram can not be shown above omg......

 

[Adequate]

I think the question is poorly worded and is supposed to be asking "horizontal velocity after 3 seconds", since it's nigh on impossible to recombine vertical and horizontal velocity back into a total velocity. So:

@ = 55 degrees
V = 120 m/s

Horizontal velocity = Vsin@ = 0.819152.120 = 98.30 m/s

But horizontal velocity is a constant, therefore the answer is B.


I_F

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er... Horizontal velocity = vcos@

not sin@

 

[insert-username]

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er... Horizontal velocity = vcos@

not sin@

Gah crap, you're dead right. That throws me right off.

EDIT: No it doesn't:

at and angle of 55 degrees from the vertical

Therefore the angle is 35 degrees from the horizontal.

Therefore the horizontal velocity = Vcos@ = 0.8191 x 120 = 98.30 m/s, so the answer is B. I think the question is poorly worded and is supposed to be asking "horizontal velocity after 3 seconds", since it's nigh on impossible to recombine vertical and horizontal velocity back into a total velocity.

but seeing as its been shot into the air and gravity which is a negative acceleration of 9.8m/s is there, shouldnt it be different to the initial velocity?especially as they included the 3sec bit.

Remember for projectile motion that horizontal velocity is a constant. In reality it's not, but for ease of calculations we assume it is. The 3 second part is to trick you. This is the only way to do it (that I know of). The question was badly worded.


I_F

 

[hsveight]

Hi.
Im new here, First post. Bit nervous.

The answers B Brother D.

We should catch up some time. You seem really nice

Peace

 

[brotha.d]

well while i was searching for help, i found a website that said this

Recombine the resulting components, if needed, to determine the object's total space motion.

and that put me off.

hsveight, thats a really cool name, u seem really funny, have my number.

 

[Ror bones]

i only read part of the question but if you have two components

horizontal and vertical velocity and you want to find the actual velocity you use pythagoras

so

velocity squared = (horizontal velocity) squared + (vertical velocity) squared

the angle can also be caculated with trig

tan thita = (horizontal velocity) / (verticat velocity)

 

[Ror bones]

ok i read the question

angle from vertical = 55

so @=90-55

@ = 35


initially

vertical velocity

Uy = U sin @

= 120 sin 35

= 68.829

horizontal velocity

Ux = U cos @

= 98.298

after 3 seconds

vertical

Vy = Uy + a t

= 68.829+(-9.8)(3)

= 39.429

horizontal

stays the same

Vx = Ux

ok so we got our two components on the velocity now we need to combine them

V^2 = Vx^2 + Vy^2

= 98.298^2 + 39.429^2

= 11217.191

V = Sqare root (11217.191)

= 105.911


so answer is c you are all wrong

haha

hope you understand the notation

 

[Riviet]

The question is badly worded. The answer could be Ror bones' or just simply the horizontal component.

 

[Ror bones]

found a nice pic

you can see that you can use pythagoras to find V if you have Vy and Vx

 

[brotha.d]

but the answer says b...

 

[Riviet]

If the answer says b, then the question is simply referring to the horizontal component of the velocity, which is 120cos35 or 98.3m/s.