Projectile motion, I cannot figure out how to do this

[deagle12]

Homework Statement

The problem is, "A footballer kicked a ball up at 50 degrees and it landed 20m away horizontaly", assuming vertical acceleration is -9.81ms^-2 how would you find the time it was in the air and the max height reached.

How would you solve this?

Homework Equations

The Attempt at a Solution

Various trigonometric functions, if you need more detail please tell me.

 

[pat666]

well you know that when it lands s_j =0
s0 s_i=20=v*cos(50) and S_j=0=vsin(50)t-9.81*.5*t^2
you now have a system of simultaneous equations... solve it
i get t=2.2s and v_init=14.11m/s
For max height.
at max height v_j=0
v_j=0^2=u^2+2as where u= vertical cmpt of v_init(14.11sin(50)) and a=-9.81,/s^2
0=(14.11sin(50))^2-2*9.81s solve
i get s=5.95m

 

[pat666]

can you let me know if this is right or useful please.

 

[GreatEscapist]

Well, first you have to have velocity, to even begin this. We know when it lands, it is going 0 m/s. So v_f= 0 m/s. Now, once you find V_i, you will need to divide that up into vertical and horizontal components of velocity. (cos * theta, sin * theta)

Once you do that, you can find the time in air with
d=v_ft + 1/2at²

You have your horizontal distance, now solve for t.
I'm not sure you could use v_f= v_i + at could work for t, because of the lack of distance. But if you did use that, t would NOT BE THE TOTAL TIME IN THE AIR. It would be the time it takes to get to the max height of the projectile. So you would have to multiply it by 2.

 

[pat666]

No GreatEscapist that is not correct. Any high school physics text will tell you that all projectiles have a final velocity that is equal to the initial horizontal component + a vertical component relative to the acceleration(assuming no air drag). use the information i gave you it is correct.

 

[GreatEscapist]

Dang it. :P
Well, I tried. Sorry about that. At least now I know I need to brush up on my projectile motion. :S

 

[ehild]

Apart from the missing t here "s_i=20=v*cos(50)" , your solution is correct. Good job !

ehild