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 = v^{2} (sin2)
/ g

= [(33.72)^{2} x sin(2 x 38.98)] / 9.81

= 113.37 m

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