. A volleyball is hit when it is 4 ft above the ground and 12 ft from a 6 ft hig

Question

. A volleyball is hit when it is 4 ft above the ground and 12 ft from a 6 ft high

     net. It leaves the point of impact with an initial velocity of 35 ft/sec at an angle

     of 27o and slips by the opposing team untouched.

   a) find the parametric equations of the trajectory (path of the ball)

   b) How high does the volleyball go and when does it reach this height? Will it

       reach the height before or after it reaches the net?

   c) Will the volleyball clear the net? If so, by how much?

   d) Where will the ball land ? If the dimension of the court are 60 ft by 30 ft and

         each side it 30 ft by 30ft, will the ball will the ball land in the court?

   e) If the net is raised to 8 ft, how high off the ground will the volleyball have to

        be in order to clear the net ( assuming all other factors remain unchanged)?

Explanation / Answer

(3) (a) Let the height of volleyball at any time t = y(t) = h + u_y* t - (1/2) g t^2 ......(1)

Horizontal Distance at any time t = x(t) = u_x* t

we know g = 32.200 ft/s^2

Launch height above ground H = 4 feet

Impact (target) elevation y(t) = 0 feet

h = H - y(t) = 4 feet

Launch speed u = 35.000 ft/s

Launch angle theta = 27 degrees

u_y = u* sin(theta) = 35*cos(27 degrees)=15.890 ft/s

u_x = u* cos(theta) = 35*sin(27 degrees)=31.185 ft/s `

Therfore fromequaition (1)

y(t) = 4 + 15.89*t - 16.1* t^2 and x(t)=31.185*t

(b) For maximium height y'(t)=0

=> 15.89-32.2t=0

=> t=15.89/32.2=0.4934 sec

therfore y(0.4934)=maximum height=7.92068 ft

(c) Time taken to travel to net 12=u_x*t => t=12/31.185=0.3848 sec

At time 0.3848 height of ball=y(0.3848)=7.73053 which is greater that height of 6 ft net

so ball will clear the net.

Gap=7.73053-6=1.73053

(d)Time of flight => y=0 =4 + 15.89*t - 16.1* t^2

solving the quadratic equation

Quadratic Equation Coefficients

A = 16.100 B = -15.890 C = -4.000

Total Flight Time seconds

From the quadratic, t = 1.19488 and t = -0.208 (since negative time is no meaning)

so t=1.19488

therofer horizontal distance=u_x=31.185*t =31.185*1.19488=37.2623 ft

so ball will hit the land in court

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