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 33° and slips by the opposing team untouched. The acceleration due to gravity is g= 32 ft/sec. Answer the following questions. a. Find a vector form for the path of the volleyball. r(t) = 35 cos (33) -ti+ 16 - 35 sin (33) - 4j (Do not evaluate. Do not include the degree symbol in your answer.)

Trigonometry (MindTap Course List)
8th Edition
ISBN:9781305652224
Author:Charles P. McKeague, Mark D. Turner
Publisher:Charles P. McKeague, Mark D. Turner
Chapter3: Radian Measure
Section3.5: Velocities
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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 33° and slips by the opposing team untouched. The acceleration due to gravity is g= 32 ft/sec. Answer the following questions.
a. Find a vector form for the path of the volleyball.
r(t) = 35 cos (33) • ti+ 16t- 35 sin (33) – 4j
(Do not evaluate. Do not include the degree symbol in your answer.)
Transcribed Image Text: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 33° and slips by the opposing team untouched. The acceleration due to gravity is g= 32 ft/sec. Answer the following questions. a. Find a vector form for the path of the volleyball. r(t) = 35 cos (33) • ti+ 16t- 35 sin (33) – 4j (Do not evaluate. Do not include the degree symbol in your answer.)
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