Physics projectile motion

[chiddler]

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If T denotes the total time of flight of the projectile, which of the following expressions correctly gives the initial vertical velocity?

a. Voy = gT/2
b. Voy = gTsin(45)
c. Voy = gT
d. Voy = 2gT

Answer: A.

Having trouble figuring this out. I know that initial position to where the projectile hits the highest peak of its path is time = T/2, but i'm not sure how to use this information to find the answer.

Thanks.

 

[chiddler]

one more question: isn't range of projectile motion equivalent to horizontal distance traveled?

 

[dmf2682]

If T denotes the total time of flight of the projectile, which of the following expressions correctly gives the initial vertical velocity?

a. Voy = gT/2
b. Voy = gTsin(45)
c. Voy = gT
d. Voy = 2gT

Answer: A.

Having trouble figuring this out. I know that initial position to where the projectile hits the highest peak of its path is time = T/2, but i'm not sure how to use this information to find the answer.

Thanks.

If initial position to peak is t /2 then the ending elevation is the same as the initial.

X= vo t - gt ^2

X is zero, solve for vo

Should get a.

 

[syoung]

one more question: isn't range of projectile motion equivalent to horizontal distance traveled?

Yes you do range = Vx(t)

 

[pfaction]

And since Vx will always stay the same, d=velocity of the original x direction * t

 

[chiddler]

If initial position to peak is t /2 then the ending elevation is the same as the initial.

X= vo t - gt ^2

X is zero, solve for vo

Should get a.

thanks that works well. can you think of a way to solve this using

a = v/t

this is what i was trying to do and it feels less like a plug n chug than using the kinematic equation you used. i'd prefer this so that i can reason it logically.

 

[chiddler]

Yes you do range = Vx(t)

frieking hell

TPR

R/2 > or < than Horizontal distance

so i panicked in trying to figure out how they are different. >.>

 

[pfaction]

a=v/t
a=dv/dt
a=vf-vi/t
at+vi=vf
rearrange it:
v=v0+at

v=vo+at
vf=0
-v0=-at
assuming g is pointing downwards
v0/a=t

vy=0 at t/2
v0 = at/2

 

[dmf2682]

thanks that works well. can you think of a way to solve this using

a = v/t

this is what i was trying to do and it feels less like a plug n chug than using the kinematic equation you used. i'd prefer this so that i can reason it logically.

I'm always hesitant to use that one because it's based off the assumption that either the starting or ending velocities is zero. Best to think of acceleration as Delta v / Delta t.

In your case you can use it.

At Max elevation vy is zero. Delta v is voy. Time to Max elevation is t/2. So g = a=voy / (t/2)

Solving gets you the same answer.

 

[chiddler]

lol i guess my approach made it a bit more complicated.

thanks very much for the great responses.

 

[chiddler]

If initial position to peak is t /2 then the ending elevation is the same as the initial.

X= vo t - gt ^2

X is zero, solve for vo

Should get a.

wait shouldn't the equation be

x=vo t + gt^2?

as opposed to vot - gt^2

edit: nevermind. gravity. sign. yeah.