Asoccer ball of mass 0.35 kg is rolling with velocity 0, 0, 1.8 m/s, when you kick it. your kick delivers an impulse of magnitude 1.8 n · s in the −x direction. the net force on the rolling ball, due to the air and the grass, is 0.26 n in the direction opposite to the direction of the ball's momentum. using a time step of 0.5 s, find the position of the ball at a time 1.5 s after you kick it, assuming that the ball is at the origin at the moment it is kicked. use the approximation vavg ≈ pf /m. (express your answer in vector form.)

position 9.58 m

Explanation:

In impulse exercises and amount of movement, we always assume that the contact time is small,

     I = Δp

With this expression we can calculate the final speed

     I = m Vf - m Vo

    Vf = (I + mVo) / m

    Vf = (1.8 + 0.35 1.8) /0.35

    Vf = 6.94 m / s

To calculate the acceleration of the ball we use Newton's second law, after finishing the impulse

     ∑ F = m a

     fr = m a

     a = fr / m

     a = -0.26 / 0.35

     a = -0.74 m/s²

A negative sign indicates that this acceleration is slowing the ball

Now we have speed and time acceleration, so we can use the kinematic equations to find the position at 1.5 s

      X = Vo t + ½ to t²

 In this case Vo is the speed with which the ball comes out after the impulse 6.94

     X = 6.94 1.5 + ½ (-0.74) 1.522

     X = 9.58 m

he kicked it 10 feet straight up into the air? i think this works.

real, smaller and upside-down.   so the answer should be b.