Magnitude and direction of its velocity problem

[Quezar]

A soccer ball is kicked with an initial speed of 5.9 m/s in a direction 24.0° above the horizontal. Find the magnitude and direction of its velocity at the following times. (Take the +x axis to be parallel to the ground, in the direction that the ball is kicked.)

(a) 0.250 s after being kicked
1 ___ m/s
2 ___ degrees (counterclockwise from the +x axis)

(b) 0.500 s after being kicked
1 ___ m/s
2 ___ degrees (counterclockwise from the +x axis)


vx = v0x +ax * t
vy= v0y + ay * t


cos24=x/5.9
1/2*-9.8*.25^2+2.4*.25 = .29
vx = 5.4 vy = .29
thats all I could figure out

 

[thebigstar25]

you are given the initial velocity and theta , you can get vox and voy noting that:

vox = vo*costheta , voy = vo*sintheta .. and in each part you given the time (t) , another thing you need to notice is that the velocity in the x-direction is constant, so what that tells you?
and the velocity in the y-direction is changing under the influence of gravity, you wrote this equation:
vy = voy -gt , you have voy , g is constant , and you are given t..

to find v it is just finding sqrt(vx^2 + vy^2) and the direction is just arctan(vy/vx) ,this question is fairly straight forward, try again and see if it is different than the values you got .. :)