Question

50. A soccer ball is kicked with an initial speed of 10.2 m/s in a direction 25.0° above the horizontal. Find the magnitude and direction of its velocity (a) 0.250 s (b) 0.500 s after being kicked. (c) Is the ball at its greatest height before or after 0.500 s? Explain.

Answer 1

We have

vi=initial velocity=10.2 m/s

θ=25°

a)

We have that for

$\begin{gathered} v_{ix}=10.2\cos (25) \\ v_{iy}=10.2\sin (25) \end{gathered}$

Then for

$v_x=v_{ix}+at$$\begin{gathered} v_x=10.2\cos (25)+(0)(0.25) \\ v_x=10.2\cos (25) \end{gathered}$

For vy

$\begin{gathered} vy=v_{iy}+(-9.8)(t) \\ v_y=10.2\sin (25)+(-9.8)(0.25)=1.86\text{ m/s} \end{gathered}$

For the magnitude of the velocity

$v=\sqrt[]{(10.2\cos 25)^2+1.86^6}=9.43\text{ m/s}$

For the direction of the velocity

$\theta=\tan ^{-1}(0.2)=11.40\text{ \degree}$

For part b)

$vx=10.2\cos (25)\text{ m/s}$

$vy=10.2\sin (25)+(-9.8)(0.5)=-0.589\text{ m/s}$

For the magnitudeof the velocity

$v=\sqrt[]{(10.2\cos (25))^2+(-0.589)^2}=9.26\text{ m/s}$

For the direction of the velocity

$\theta=\tan ^{-1}(\frac{-0.589}{10.2\cos (25)})=-3.65\text{ \degree}$

For c)

Because of the negative sign that we obtain in the velocity, we can say that the greatest height is located before the 0.5 m