A soccer player kicks the ball toward a goal that is 23.0 m in front of him. The ball leaves his foot at a speed of 17.6 m/s and an angle of 29.0 ° above the ground. Find the speed of the ball when the goalie catches it in front of the net.

Answer 1

Vo = 17.6m/s[29o]
Xo = 17.6*Cos29 = 15.4 m/s
Yo = 17.6*sin29 = 8.53 m/s

Y = Yo + g*Tr = 0
Tr = -Yo/g = -8.53/-9.8 = 0.870 s. = Rise time.

Range = Dx = Vo^2*sin(2A)/g =
17.6^2*sin(58)/9.8 = 26.8 m.

h = Yo*Tr + 0.5g*Tr^2 =
8.53*0.870 - 4.9*0.870^2 = 3.71 m. = Max. ht.

Xo*(Tr+Tf) = 23 m
15.4 *(0.87+Tf) = 23
13.4 + 15.4Tf = 23
15.4Tf = 23-13.4 = 9.6
Tf = 0.624 s.

h = ho - 0.5g*Tf1^2 =
3.71 - 4.9*0.624^2 = 1.80 m. = ht. of the ball when it reaches the net.

Y^2 = Yo^2 + 2g*h = 0 + 19.6*(3.71-1.8)=
37.4
Y = 6.12 m/s = Ver. component of velocity when the ball reaches the net.

V = Xo + Yi = 15.4 + 6.12i
V^2 = 15.4^2 + 6.12^2 = 274.61
V = 16.6 m/s = Total velocity when caught by the goalie.

Note: Since the range is greater than the distance from the kicker to the goal, the ball(if not touched) will land
26.8 meters from the kicker which is 3.8
behind the net.