A regulation volleyball court is L = 18.0 m long and a regulation volleyball net

Question

A regulation volleyball court is L = 18.0 m long and a regulation volleyball net is d = 2.43 m high. A volleyball player strikes the ball a height h = 1.97 m directly above the back line, and the ball's initial velocity makes an angle theta = 56 degree with respect to the ground (see the figure). At what initial speed must the ball be hit so that it just barely makes it over the net? (Assume the volleyball is hit so that its path is parallel to the side-line as seen from an observer directly above the court, and that the volleyball is a point object.) At what initial speed must the ball be hit so that it lands directly on the opponent's back line?

Explanation / Answer

12) Vertical distance travelled to clear the net is d-h = 2.43-1.97 = 0.46 m

using kinematic equations

vertical direction

y =(Voy*t) + (0.5*ay*t^2)

Voy is the y-component of initial velocity = Vo*sin(56) = 0.83*Vo

ay is the vertical component of accelaration

Vo is the required initial speed

0.46 = (Vo*0.83*t)-(0.5*9.8*t^2)

Horizontal direction

x = (Vo*cos(56))*t

(18/2) = (Vo*0.56)*t

t= 9/(Vo*0.56) = 16.1/Vo

then

0.46 = (Vo*0.83*(16.1/Vo)) - (0.5*9.8*(16.1/Vo)^2)

Vo = 9.92 m/sec

13) x = 2L = 2*18 = 36 m

but X = Vo*cos(56)*t

36 = 0.56*Vo*t

t = 36/(0.56*Vo) = 64.3/Vo

and also using

y = (Vo*sin(56)*t) +(0.5*ay*t^2)

-(2.43-1.97) = (Vo*0.83*t) -(0.5*9.8*t^2)

-0.46 = (Vo*0.83*(64.3/Vo)) - (4.9*(64.3/Vo)^2)

Vo = 19.4 m/sec

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