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P Purdue Web Authentication X W Quiz1 Search Q&A; I Chegg.com X e Connecting X CAM #1 HP Prime Graph... X https://www. t/web/Student/Assig 3975863 Search 2. C+ -l 5 points YF14 3.P018 above the horizontal. The shot hits the ground 2.18 s later. You can ignore air resistance. (As A shot putter releases the shot some distance above the level ground with a velocity of 10.2 m/s, 53.0 direction of the shot and upward are positive.) What are the components of the shot's acceleration while in flight? m/s2 (horizontal component) m/s2 (vertical component) (b) What are the components of the shot's velocity at the beginning of its trajectory? m/s (horizontal component m/s (vertical component) What are the components of the shot's velocity at the end of its trajectory? m/s (horizontal component m/s (vertical component) (c) How far did she throw the shot horizontally? Why d sin 2ao O The initial and final heights are different. O It is not a projectile motion. O There is air drag. How high was the shot above the ground when she released it? Draw x-t, y-t, vx t and vy t graphs for the motion. (Do this on paper Your instructor may ask you to turn in this work.) Submit Answer Save Progress I'm Cortana. Ask me anything the horizontal 6:21 AM ESP 6/23/2016

Explanation / Answer

a)

horizontal acceleration is 0
vertical acceleration is -9.81m/s^2 due to gravity

b)

10.2sin(53) = 8.146 m/s vertical
10.2cos(53) = 6.1385 m/s horizontal

8.146 - (9.81)(2.18) = -13.2398 m/s vertical
6.1385 m/s = horizontal

c)

d = (6.1385)(2.18) = 13.3819 m

e)

a = -9.81ms^-2
u = 8.146
t = 2.18 s
s = ?
You then put these value into the kinimatic equation:
s = ut + 0.5a(t^2)

s = (8.146 x 2.18) - (1/2 x 9.81 x 2.18^2)

s = 17.7583 - 23.31

s = 5.55 m

You will get a negative value for s, but this is correct because the ground is that far under the point of throwing.

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