Question

You kick a soccer ball with a force (F) of 1200 N at an angle of 45 degrees. The soccer ball has an initial velocity of (v i ) of 5 m/s at 0 degrees , a mass (m) of 0.4 kg, and the time of contact (t) between yo ur foot and the ball is 0.01 s. • H ow far does the ball travel before it hits the ground (∆s x ) ? • Express your initial velocity in the projectile phase with polar coordinates (r, θ)

Answer 1

Force = dP/dt
Where dP is the change in momentum and dt is the contact time.
dP = Fdt
= 1200 x 0.01 = 12 kg m/s
dP = mdv, where m is the mass of the ball.
dv = dp/m
= 12/0.4
= 30 m/s

So the velocity added to the ball is 30 m/s at 45 degrees. The ball already had a velocity of 5 m/s
Net velocity along horizontal direction, vx = 30 cos(45) + 5 = 26.21 m/s
Net velocity along vertical direction, vy = 30 sin(45) = 21.21 m/s
Net velocity, v = SQRT[(vx)² + (vy)²]
= 33.72 m/s
tan = 21.21/26.21
= tan^-1(21.21/26.21)
= 38.98 degrees.
v (r, ) = (33.72, 38.98)

Range of the ball, R = v² (sin2) / g
= [(33.72)² x sin(2 x 38.98)] / 9.81
= 113.37 m