Question

In a World Cup soccer match, Juan is running due north toward the goal with a speed of 7.50 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 32,9° east of north, relative to the ground.Part AWhat is the magnitude of the ball's velocity relative to Juan?Part BWhat is the direction of the ball's velocity relative to Juan?

Answer 1

Answer:

a) The magnitud will be $7.75\frac{m}{s}$.

b) The direction of the ball's velocity relative to Juan will be $\phi = 25.44^\circ$.

Explanation:

$\theta = 32.9^\circ$

Juan's velocity relative to the ground : $v_{j,g}=(0 ; 7.5)\frac{m}{s}$

Ball's velocity relative to the ground : $v_{b,g}=(12.9sin(\theta ) ; 12.9cos(\theta ))\frac{m}{s}$

We know that Ball's velocity relative to the ground is Ball's velocity relative to Juan plus Juan's velocity relative to the ground. This is a very important notion that even extends to more complex mathematical problems (differentiation).

$v_{b,g}=v_{b,j} + v_{j,g}$ ⇒ $v_{b,j} =v_{b,g}-v_{j,g}$

⇒ $v_{b,j}= (12.9sin(\theta ) ; 12.9cos(\theta ))\frac{m}{s}  - (0 ; 7.5)\frac{m}{s} = [tex](7 ; 10.83)\frac{m}{s}  - (0 ; 7.5)\frac{m}{s}$[/tex]

∴ $v_{b,j}= (7 ; 3.33)\frac{m}{s}$

a) The magnitud will be $\left | v_{b,j}  \right |=\sqrt{7^2 + 3.33^2} =7.75\frac{m}{s}$.

b) The direction of the ball's velocity relative to Juan will be:

$tan(\phi )=\frac{3.33}{7}$ ⇒ $\phi =arctan(\frac{3.33}{7})$

∴ $\phi = 25.44^\circ$.