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

Answer:

Part a)

$v_{bj} = 11.03 m/s$

Part b)

$\theta = 4.57 degree$ East of South

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

$v_{bj} = \sqrt{11^2 + 0.88^2}$

$v_{bj} = 11.03 m/s$

Part b)

direction of velocity of the ball

$tan\theta = \frac{v_y}{v_x}$

$tan\theta = \frac{-0.88}{11}$

$\theta = 4.57 degree$ East of South