keTutonialAssignment.as Home I Student l.com Class Management I Help Homework Ch

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keTutonialAssignment.as Home I Student l.com Class Management I Help Homework Ch.2 Begin Date: 9/4/2018 12:01:00 AM- Due Date: 9/18/2018 11:59.00 PM End Date: 9/18/2018 11:59.00 PM (8%) Problem S: A rock is thrown straight down from the Verrazano Narrows Bridge in New York City with an antal speed of 12.5 m/s. The roadway of this bridge is 70 m above the water. Take upwards to be the positive direction #13% Part (a) Calculate the displacement at the time of 10 1.0)-13 Incorrect! > & 13% Part (b) Calculate the velocity at the time of 1 0 Grade Sam Potechial cosO cotanOasin acos0 sino 4 5 6 1 23 Amempts remaiu per aamp atan) acotan coshO tanb cotanhO ° Degrees O Radians Hist I give up Hints: 1 dedaction per hist. Hists remainieg Feedback: deduction per Beedback 13% Part (c) Calculate the displacement at the time of 1.5 s 13% Part (d) Calculate the velocity at the time of 1.5 s 13% Part (e) Calculate the displacement at the time of 2.0 s. 13% Part (f) Calculate the velocity at the time of 2 s 13% Part Cg) Calculate the displacement at the tune of 2.5 s. 13% Part (h) ii - d Calculate the velocity at the tate of 2.5 s. DOLL

Explanation / Answer

We know that displacement is given by:

d = u*t + 0.5*a*t^2

final speed is given by:

v = u + a*t

Given that

u = -13.5 m/sec^2

a = -g = -9.8 m/sec^2

Part A.

At t = 1.0 sec

d = -13.5*1 - 0.5*9.8*1^2

d = -18.4 m

Part B.

At t = 1.0 sec

v = -13.5 - 9.8*1

v = -23.3 m/sec

Part C.

At t = 1.5 sec

d = -13.5*1.5 - 0.5*9.8*1.5^2

d = -31.275 m

Part D.

At t = 1.5 sec

v = -13.5 - 9.8*1.5

v = -28.2 m/sec

Part E.

At t = 2.0 sec

d = -13.5*2.0 - 0.5*9.8*2.0^2

d = -46.6 m

Part F.

At t = 2.0 sec

v = -13.5 - 9.8*2

v = -33.1 m/sec

Part G.

At t = 2.5 sec

d = -13.5*2.5 - 0.5*9.8*2.5^2

d = -64.375 m

Part H.

At t = 2.5 sec

v = -13.5 - 9.8*2.5

v = -38 m/sec

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