Question

A man kicks the 150-g ball such that it leaves the ground at an angle of 60° and strikes the ground at the same elevation a distance of 12 m away (14 points). Determine: (a). The velocity of the ball instantly after the man kicks the ball; (6 points) (b). Determine the impulse of his foot on the ball at A. (8 points) Neglect the impulse caused by the ball’s weight while it’s being kicked.

Answer 1

Answer:

$313.92\ \text{m/s}$

$47.088\ \text{kg m/s}$

Explanation:

m = Mass of ball = 150 g

$\theta$ = Angle of kick = $60^{\circ}$

$x$ = Displacement of ball in x direction = 12 m

Range of projectile is given by

$x=\dfrac{u\sin^2\theta}{2g}\\\Rightarrow u=\dfrac{2xg}{\sin^2\theta}\\\Rightarrow u=\dfrac{2\times 12\times 9.81}{\sin^260^{\circ}}\\\Rightarrow u=313.92\ \text{m/s}$

The velocity of the ball instantly after the man kicks the ball is $313.92\ \text{m/s}$

Impulse is given by

$J=mu\\\Rightarrow J=0.15\times 313.92\\\Rightarrow J=47.088\ \text{kg m/s}$

The impulse of his foot on the ball is $47.088\ \text{kg m/s}$.