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zloy xaker [14]
3 years ago
5

Eddie the eagle, british olympic ski jumper, is attempting his most mediocre jump yet. after leaving the end of the ski ramp, he

lands downhill at a point that is displaced 72.1 m horizontally from the edge of the ramp. his velocity just before landing is 33.0 m/s and points in a direction 30.0$^\circ$ below the horizontal. neglect any effects due to air resistance or lift. what was the magnitude of eddie's initial velocity as he left the ramp?
Physics
1 answer:
DedPeter [7]3 years ago
5 0
When he lands his horizontal velocity is 
<span>28 cos40 = 21.45 m/s </span>

<span>the time in flight comes from </span>
<span>x = Vht </span>
<span>58.8 = 21.45t </span>
<span>t = 2.74 seconds </span>

<span>his vertical velocity at landing is </span>
<span>28 sin40 = -18 m/s </span>

<span>his vertical velocity equation is </span>
<span>v = V0 - gt </span>
<span>-18 = V0 - 9.81(2.74) </span>
<span>V0 = -18 + 9.81(2.74) </span>
<span>V0 = 8.88 </span>

<span>his velocity magnitude was </span>
<span>v = (8.88^2 + 21.45^2)^½ </span>
<span>v = 23.2 m/s ANSWER </span>

<span>his initial direction was </span>
<span>tanθ = 8.88/21.45 </span>
<span>θ = 22.5 degrees above the horizontal ANSWER </span>

<span>to find the time to the flight apex from launch </span>
<span>v = gt </span>
<span>8.88 = 9.81t </span>
<span>t = 0.905 s </span>

<span>in 0.905 s Eddie has risen how far above the edge </span>
<span>y = ½(9.81)(0.905^2) </span>
<span>y = 4 m </span>

<span>the remainder of the flight is all drop and takes 2.74 - 0.905 = 1.85 seconds </span>

<span>in 1.85 seconds he drops </span>
<span>y = ½(9.81)(1.85^2) </span>
<span>y = 16.7 m </span>

<span>so the height from the edge to the landing point is </span>
<span>16.7 - 4 = 12.7 m ANSWER</span>
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Complete Question

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The correct option is B

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Brandon buys a new Seadoo. He goes 22 km north from the beach. He then goes 11km to the east. Then chases a boat 3 km north. Wha
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A ___ force is not path-dependent and does not change the overall mechanical energy of an object.​
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Answer:

conservative

Explanation:

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hope this helped!

6 0
3 years ago
Read 2 more answers
PLEASE HELP LOTTA POINTS
sergiy2304 [10]

There's a nasty wrinkle here that's kind of sneaky, and makes the work harder than it should be.

Look at the first question.  There's a number there that's dropped in so quietly that you're almost sure to miss it, but it changes the whole landscape of both of these problems.   That's where it says

" ... 20 cm mark (30 cm from the fulcrum) ... " .

That tells us that the yellow bar resting on the pivot is actually a meter stick, but the pictures don't show the centimeter marks on the stick.  The left end of the stick is "0 cm", the right end of the stick is "100 cm", and the pivot is under the "50 cm" mark.  

When the question talks about hanging a weight, it tells the <em>centimeter mark on the stick</em> where the weight is tied.  To solve the problem, we have to first figure out <em>how far that is from the pivot</em>, then calculate how far from the pivot to put the weight on the other side, and finally <u><em>what centimeter mark that is</em></u> on the stick.      

How to solve the problems:

-- The "moment" of a weight is (the weight) x (its distance from the pivot) .

-- To balance the stick, (the sum of the moments on one side) = (the sum of the moments on the other side).

= = = = = = = = = =  

#1).  Only one moment on the left side.  

(160 gm) x (30 cm from pivot) = 4,800 gm-cm

To balance, we need 4,800 gm-cm of moment on the right side.

(500 gm) x (distance from pivot) = 4,800 gm-cm

Distance from pivot = (4,800 gm-cm) / (500 gm)  =  9.6 cm

The 500 gm has to hang 9.6 cm to the right of the pivot.  But that's not the answer to the problem.  They want to know what mark on the stick to hang it from.  The pivot is at the 50cm mark.  The 500gm has to hang 9.6 cm to the right of the pivot.  That's the <em>59.6 cm</em> mark on the stick.

= = = = =

#2).  There are 2 weights hanging from the left side. We have to find the moment of each weight, add them up, then create the same amount of moment on the right side.

one weight:  120gm, hanging from the 25cm mark.

That's 25cm from the pivot.  Moment = (120gm) (25cm) = 3,000 gm-cm

the other weight:  20gm, hanging from the 10cm mark;

That's 40cm from the pivot.  Moment = (20gm) (40cm) = 800 gm-cm

Add up the moments on the left side:

(3,000 gm-cm) + (800 gm-cm) = 3,800 gm-cm.

To balance, we need 3,800 gm-cm of moment on the right side.

(500 gm) x (its distance from the pivot) = 3,800 gm-cm

Distance from the pivot = (3,800 gm-cm) / (500 gm) = 7.6 cm

The pivot is at the 50cm mark on the stick.  You have to hang the 500gm from 7.6cm to the right of that.  The mark at that spot on the stick is                (50cm + 7.6cm) = <em>57.6 cm </em>.

3 0
3 years ago
Read 2 more answers
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