A2 kg ball with an initial velocity of 10 m/s moves at an angle 60º above the +x-direction. the ball hits a vertical wall and bounces off so that it is moving 60º above the −x-direction with the same speed.

what is the impulse delivered by the wall?

I = 20 i ^ N s

Explanation:

For this problem let's use the Impulse equation

       I = Δp = m - v₀

The impulse and the velocity are vector quantities, let's calculate on each axis, let's decompose the velocity

     cos 60 = vₓ / v

     vₓ = v cos 60

     sin60 = / v

      = v sin60

     vₓ = 10 cos 60

      = 10 sin60

    vₓ = 5.0 m / s

    = 8.66 m / s

Let's calculate the impulse on each axis

X axis

     Iₓ = m - m vₓ₀

How the ball bounces

    = - vₓ₀ = vₓ

    Iₓ = 2 m vₓ

    Iₓ = 2 2 5

    Iₓ = 20 N s

Y axis

    = m - m vyo

On the axis and the ball does not change direction so

    = vyo

  = 0

The total momentum is

   I = Iₓ i ^ + j ^

   I = 20 i ^ N s

the results is a decrease in air temperature with an increase of altitude.

explanation:

the answer is a 3.8cm

explanation: