11:15 a Google Bil Search TextD HW 2-2D+ Motion Begin Date: 8/31/2018 12:00 18 1

Question

11:15 a Google Bil Search TextD HW 2-2D+ Motion Begin Date: 8/31/2018 12:00 18 12:00.00 AM-Due Date: 9/9/2018 11 :59.00 PM End Date:919/2018 11:59.00 (6%) Problem 10: of 60° above the horizontal. It lands on the top edge of the cliff 3.85 s later. An arow is shotfrom a height of 1.8m to and aclfrofheight lihis shot witível atyerm sank a 33% Part (a) what is the height H or the cliff in m? 393 Cerrect 33% Part (b) what is the maximum height reached by the arrow along its tri ectory in meters? cotan) asin acost0 atano acotan) sih per anenpt 1 2 3 detailed v °Degrees Radians I give up Hinh: 0% hint Hinda remaining tion 33 % Part (c) what is the arrow's speed just before hitting the cliff in mo? 2 4 5 6 8

Explanation / Answer

As we know that;

= Y = vertical displacement = final height - initial height

= H - 1.8

Where,

= ay = acceleration in Y-direction = - 9.8 m/s^2

= t = time taken = 3.85 sec

= Voy= initial velocity in Y-direction

= 33 Sin60 = 28.57 m/s

Then, Using the equation,

= Y = Voyt + (0.5)ayt^2

= H - 1.8 = 28.57 (3.85) + (0.5) (-9.8) (3.85)^2

= H = 39.16 m

Then,

Final velocity in vertical direction is given as,

= Vfy = Voy+ ayt

= Vfy= 28.57 + (-9.8) (3.85)

= Vfy= - 9.16 m/s

So, along the X-direction :final speed

= Vfx= Vox= Vo Cos60

= 33 Cos60 = 16.5 m/s

Then,

Net speed,

= V = sqrt (Vfx^2 + Vfy^2)

= sqrt (16.5^2 + (-9.16)^2) = 18.87 m/s

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