2D Vector Kinematics Problem (projectile)

[j3llzang]

Homework Statement

A person kicks a ball h of 2.0 m above the floor. If the ball flies at some angle and lands 1.2 s later at horiz d of 6.5 m from its orig. place, calc. the orig. horiz and vert. comp.s of the ball's v.
Calc. the final components of the ball's v.
(Assume the ball has no horiz. accel.)

Homework Equations

v = vo + at
v^2 = vo^2 + 2ad
d = vot + 1/2 (at^2)
g = -9.8 m/s^2

The Attempt at a Solution

Although I am given the time, I cannot divide time into 2 for getting the time it takes to reach the highest point (since it has to come down the 2.0m height)
I found out vx (which is constant throughout) = (6.5m / 1.2s) = 5.4167 m/s
But I am stuck.
Any idea? (or is there not enough info given in the question?)

Cheers,

 

[Sourabh N]

to find the vertical component, simply use the second equation of motion (just write down the equation and you will realize that you have enough info). Don't worry about the going up and coming down issue, the equation takes care of it :-)

 

[j3llzang]

to find the vertical component, simply use the second equation of motion (just write down the equation and you will realize that you have enough info). Don't worry about the going up and coming down issue, the equation takes care of it :-)

You mean v^2 = vo^2 + 2ad?
or
d=vot + 1/2 (at^2)

Even if I were to use either one, I can't solve vf or vo for vertical, since I do not have dy (it would be bigger than 2.0m, because it moves up first)
I am also not given the vo...
thanks for the quick reply, though :)

 

[Sourabh N]

Yes the second equation you wrote is the one useful here.

And remember, d is the vertical "displacement", not distance. So whatever way the ball moved (up or down), it's the final position that matters.

Imagine I throws a projectile from ground. After it comes back to ground doing the parabolic motion, it's vertical displacement is.. ZERO!

 

[j3llzang]

Yes the second equation you wrote is the one useful here.

And remember, d is the vertical "displacement", not distance. So whatever way the ball moved (up or down), it's the final position that matters.

Imagine I throws a projectile from ground. After it comes back to ground doing the parabolic motion, it's vertical displacement is.. ZERO!

Ahh,, I see your point! :)
However, there's still a problem of finding v_oy or v_fy...
I can't put any of these two as zero, since there is the initial "kick" to the ball, which means at the bottom of its parabolic shape (above the 2.0m height), its v is not going to be zero...
Any idea? :P

 

[Sourabh N]

I"m not sure what you mean (which is probably my fault), so I'll assume what I think is what you mean and answer.

The second equation you wrote in Post #3, should give you initial vertical velocity component (since you know d, t and g). Then the first equation in same post should give you final vertical velocity component (since you know everything else).

Oh, and you know the final horizontal velocity already!

 

[j3llzang]

aha! I misinterpreted your reply.
I got it all figured out now. Thanks a lot ;)