Projectile

[Format]

Need help with this one:

A Projectile is launched at 35.0 degrees above the horizontal with an intial velocity of 120m/s. What is the projectiles speed 3.00 seconds later?

Ive got the x and y compontents...wut now?

[faust9]

This is what you have:
\vec v=v_xi+v_yj

Think of the above like a right triangle with the speed as the hypotenuse. How would you go about finding the length if this were a trig or algebra class?

[UrbanXrisis]

The projectial's horizontal speed would be the x component
Find the vertial speed by using

a=-9.8m/s^2
t=3s
SpeedInital=y-component
SpeedFinal=solve for this

[Format]

well i tried using Vf = Vi + at ...which has all i need but it doesnt give the correct answer (106m/s)

[Format]

ohh, nevermind i get it...its the resultant of the 2 components :biggrin: thx

[faust9]

well i tried using Vf = Vi + at ...which has all i need but it doesnt give the correct answer (106m/s)

It wont give you the correct answer. The speed of the projectile is the magnitude of the velocity vector. Velocity vector has two components--velocity in the x direction and velocity in the y direction. You said you found the x and y components already so you have your vector components.

[faust9]

ohh, nevermind i get it...its the resultant of the 2 components :biggrin: thx

Good. You figured it out before I finished typing.

[Format]

lol yea thx! One other thing concerning projectiles...how do you solve for maximum height with only an initial velolcity at a certain angle?

[Pyrrhus]

At Max Height, Vy will be 0, you know the vector is tangent to the parabole, if you know your calculus, you know the vector will be horizontal at max height.

[Format]

hmm...still not gettin it

[Pyrrhus]

Show your work.

[Format]

Y Components:
Vi = 7.89
Vf = 0
a = -9.8
t = 1.62
d = ?

d = Vit + .5(a)t^2
0 = 7.89t + .5(-9.8)t^2
t = 1.62

d = 7.89(1.62) + .5(-9.8)(1.62)^2

better?

[Format]

lol nevermind im dumb...i got it