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A physics professor did daredevil stunts in his spare time. His

last stunt was an attempt to jump across a river on a motorcycle

(the figure (Figure 1) ). The takeoff ramp was inclined at 53.0 ?,

the river was 40.0 m wide, and the far bank was 15.0 m lower than

the top of the ramp. The river itself was 100 m below the ramp. You

can ignore air resistance.

A) What should his speed have been at the top of the ramp for

him to have just made it to the edge of the far bank?

B) If his speed was only half the value found in A, where did he

land?

Edit
Added Sun, 03 Jul '16
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A) taking the horizontal components (x) u will get

:

Vox = 40 / t

=> Vo cos (53) = 40 / t

so Vo = 66.46 / t .... (1)

hence Vo * t = 66.46

________________________

taking the vertical component (y) and applying

:

d = Vo t 1/2 a t^2 u will get :

- 15 = Vo sin53 t - 1/2 g t^2

so - 15 = 0.8 Vo t - 4.9 t^2

replacing Vo t by 66.46 from eq(1) you will get

:

- 15 - 53.17 = - 4.9 t^2

so t = 3.73 s

hence Vo = 66.46 / 3.73 = 17.81 m/s

_______________________________________________________________

B) assuming horizontal distance x :

y = Voy t - 1/2 g t^2

- 100 = 8.9 t - 4.9 t^2

so t = 5.5 seconds

so

Vox = 8.9 cos53 = x / t

hence x = 8.9 cos53 * 5.5

x = 29.46 m from the ramp

Edit
Added Sun, 03 Jul '16
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My Answere to part B should be correct:

If his speed was only half the value found in A, where did he land?

v = ½ * 17.83115661 = 8.915578305 m/s

Horizontal = v * cos 53, Vertical = v * sin 53

Let’s determine the initial horizontal and vertical velocity.

Horizontal = 8.915578305 * cos 53

Vertical = 8.915578305 * sin 53

Let’s use the equation below to determine the time for the professor to move from the ramp to the river,

d = v * sin 53 * t – ½ * 9.8 * t^2

d = -100

-100 = 8.915578305 * sin 53 * t – 4.9 * t^2

4.9 * t – 8.915578305 * sin 53 * t – 100 = 0

Let’s solve this quadratic equation to determine the time.

I use the following website to solve quadratic equations.

http://www.math.com/students/calculators...

t = 5.302154267371364 seconds

Let’s use this time and the initial horizontal velocity to determine the horizontal distance the professor travels.

Initial horizontal velocity = 8.915578305 * cos 53

Horizontal distance = 8.915578305 * cos 53 * 5.302154267371364 = 28.44886229 meters

Since this is less than 40 meters, the professor lands in the river!

Edit
Added Tue, 26 Sep '17
I found the answere to this from this website: https://answers.yahoo.com/question/index?qid=20140130123447AAH7e9O Member 337 Tue, 26 Sep '17
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