Question

In a World Cup soccer match, Juan is running due north toward the goal with a speed of 7.60 m/s relative to the ground. A teammate passes the ball to him. The ball has a speed of 12.9 m/s and is moving in a direction of 31.4° east of north, relative to the ground.a. What is the magnitude of the ball's velocity relative to Juan?b. What is the direction of the ball's velocity relative to Juan?

Answer 1

Part a)

`v_(bj) = 11.03 m/s`

Part b)

`\theta = 4.57 degree` East of South

Step-by-step explanation:

Part a)

Velocity of Juan is given as

`v_1 = 7.60 m/s \hat j`

velocity of the ball is given as

`v_2 = 12.9(cos31.4 \hat i + sin31.4\hat j)`

now we have

`v_2 = 11\hat i + 6.72\hat j`

Part a)

We need to find velocity of ball with respect to Juan

so it is given as

`v_(bJ) = \vec v_b - \vec v_j`

`v_(bj) = 11\hat i + 6.72 \hat j - 7.6\hat j`

`v_(bj) = 11\hat i - 0.88\hat j`

magnitude of the speed is given as

`v_(bj) = √(11^2 + 0.88^2)`

`v_(bj) = 11.03 m/s`

Part b)

direction of velocity of the ball

`tan\theta = (v_y)/(v_x)`

`tan\theta = (-0.88)/(11)`

`\theta = 4.57 degree` East of South