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

Question

A volleyball hit when it is 4ft above the ground and 12 ft from a 6-ft-high net. It leaves the point of impact with an initial velocity of 44 ft/sec at an angle of 32 degreesand slips by the opposing team untouched. The acceleration due to gravity is g=32 ft/sec^2. Answer the following questions.

a. Find a vector form for the path of the volleyball.

b. How high does the volleyball go, and when does it reach maximum height?

c. Find the range and flight time.

d. When is the volleyball 7 ft above the ground? How far (ground distance) is the volleyball from where it will land?

e. Suppose that the net is raised to 8 ft. Will the ball still clear the net?

Explanation / Answer

Solution:

Height at any time t = y(t) = h + Uy t - 1/2 g t^2
Distance at any time t = x(t) = Ux t
g = 32 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 = 44 ft/s
Launch angle theta = 32 degrees
Uy = U sin(theta) = 23.32 ft/s
Ux = U cos(theta) = 37.31 ft/s `

Solve for T in: y(T) = 4 + 23.32 T - 16 T^2
Quadratic Equation Coefficients
A = 16 B = - 23.32 C = - 4

Total Flight Time seconds
From the quadratic, T = 1.61, T = -0.155

Time to reach max height seconds
tmax = Uy/g = 23.32/32 = 0.729 <================ ANS

Max Height above h feet
Y(tmax) = h + Uy tmax - 1/2 g tmax^2 = 4 + 23.32*0.729 - 1/2 * 32 * 0.729^2 = 4 + 17 - 8.50 = 12.50

Range at Max Height feet
X(tmax) = Ux * Uy/g = 37.31 * 23.32/32 = 27.19

Max Range to Impact feet
x(T) = Ux * T = 37.31 * 1.61 = 60.07

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