Answer:
[tex]313.92\ \text{m/s}[/tex]
[tex]47.088\ \text{kg m/s}[/tex]
Explanation:
m = Mass of ball = 150 g
[tex]\theta[/tex] = Angle of kick = [tex]60^{\circ}[/tex]
[tex]x[/tex] = Displacement of ball in x direction = 12 m
Range of projectile is given by
[tex]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}[/tex]
The velocity of the ball instantly after the man kicks the ball is [tex]313.92\ \text{m/s}[/tex]
Impulse is given by
[tex]J=mu\\\Rightarrow J=0.15\times 313.92\\\Rightarrow J=47.088\ \text{kg m/s}[/tex]
The impulse of his foot on the ball is [tex]47.088\ \text{kg m/s}[/tex].
