Question

Volleyball 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 27° and slips by the opposing team untouched.

a. Find a vector equation for the path of the volleyball.
b. How high does the volleyball go, and when does it reach maximum height?
c. Find its 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. Does this change things? Explain. 38. Shot put In

Answer 1

Answer:

a) the vector equation is r(x,y) = (35cos27°t, 16t^2 -35sin27° - 4)

b) Maximum height = 7.945ft

c) Time of flight = 1.201 secs and range = 37.45ft

d) when t = 0.74, the distance above the ground = 14.37ft

When t = 0.254, the distance above the ground = 26.53ft

e) if the ball is raised to 8ft, the ball won't be able to reach it. This is because the maximum height is 7.954ft

Step-by-step explanation:

a) From the diagram

x0 = 0

y0= 4

V0(initial velocity) = 35

α= 27°

y = 4 + 35sin27°t - 16t^2

y = -16t^2 + 35sin27°t+ 4

y = 16t^2 - 35sin27°t - 4

x = 35cos27°t

Vector equation for the path of the volleyball = r(x,y)

r(x,y) = (35cos27°t, 16t^2 -35sin27° - 4)

b) the ball reaches maximum when dy/dt = 0

y = 16t^2 - 35sin27°t - 4

dy/dt = 32 - 35sin27°

0 = 32 - 35sin27°t

-32 = -35sin27°t

t = -32 / -35sin27°

t = 0.4966 seconds

Put t = 0.4966 into y = 16t^2 - 35sin27°t - 4

y = 16(0.4966)^2 - 35sin27°(0.4966) - 4

y = 7.945 ft

The maximum height is 7.945 ft

c) To find the range and flight time, put y= 0

0 = 16t^2 - 35sin27°t - 4

0 = 16t^2 - 15.89t - 4

Using quadratic equation general formula,

[-b +/- √b^2 -4ac] / 2a

a = 16, b = -15.89, c = -4

= [-(-15.89) +/- √ (-15.89)^2 - 4(16)(-4)] /2(16)

= [15.89 +/- √(15.89)^2 + 256] / 32

= (15.59 +/- 22.55) / 32

= (15.89 + 22.55) / 32 or (15.89 - 22.55) / 32

= 1.201 or -0.208

Time of flight = 1.201 secs

Range = x = 35cos27°t

Range = 32cos27°(1.201)

= 37.45 ft

d) when the volley is 7ft above ground, y = 7

Recall that y = 16t^2 - 35sin27°t - 4

7 = 16t^2 - 35sin27°t - 4

0 = 16t^2 - 35sin27°t - 4 +7

0 = 16t^2 - 35sin27°t + 3

0 = 16t^2 - 15.89t + 3

Using quadratic equation general formula,

[-b +/- √b^2 -4ac] / 2a

a = 16, b = -15.89, c = 3

= [-(-15.89) +/- √ (-15.89)^2 - 4(16)(3)] /2(16)

= [15.89 +/- √(15.89)^2 - 192] / 32

= (15.59 +/- 7.7778) / 32

= (15.89 + 7.7778) / 32 or (15.89 - 7.7778) / 32

= 0.74 or 0.254

When t = 0.74,

x = 35cos27°t

x = 35cos27°(0.74)

x = 23.08 ft

Therefore , R - x(0.74)

= 37.45 - 23.08

= 14.37ft

When t = 0.254,

x = 35cos27°t

x = 35cos27°(0.254)

x = 7.92 ft

Therefore R - x(0.254) =

37.45 - 7.92 = 29.53ft

e) if the ball is raised to 8ft, the ball won't be able to reach it. This is because the maximum height is 7.954ft