c x Mail- Moses, Marcus S..... x Hw.Ch4.dvi. HW-Ch4.pdf x 3 Homework 4-Projectil

Question

c x Mail- Moses, Marcus S..... x Hw.Ch4.dvi. HW-Ch4.pdf x 3 Homework 4-Projectile M... x A Firefighter A Distance D 3 www.webassign.net/web/Student/Assignment-Responses submit?dep-13131533 the base of the counter. (a) With what velocity did the mug leave the counte 3.52 m/s (b) What was the direction of the mug's velocity just before it hit the floor? 54.90 o (below the horizontal) Need Help? Read it Master It Talk to a Tutor -11 points SerPSE8 Soln My Notes Ask Your Teacher A projectile is fired in such a way that its horizontal range is equal to 5.5 times its maximum height. What is the angle of projection? Need Help? Read it Talk to a Tutor Submit Answer Save Progress Practice Another Version 6. C+ 2 points SerPSE8 4. P013. My Notes Ask Your Teacher Chinook salmon are able to move through water especially fast by jumping out of the water periodically. This behavior is called porpoising. Suppose a salmon swimming in still water jumps out of the water with velocity 6.47 m/s at 46.9 above the horizontal, sails through the air a distance L before returning to the water, and then swims the same distance L underwater in a straight, horizontal line with velocity 3.76 m/s before jumping out again. a) Determine the average velocity of the fish for the entire process of jumping and swimming underwater. m/s (b) Consider the time interval required to travel the entire distance of 2L. By W hat percentage is this time interval reduced by the jumping/swimming process compared with simply swimming underwater a 3.76 m/s ss 2/11/2016 2:52 PM

Explanation / Answer

. 5) This is a missile launch.

We seek the maximum height

Vy2= Voy2– 2 g Y Vy = 0 0 = Voy2– 2 g Ymax

Ymax = Voy2/ 2g

We calculate the flight time

Vy = Voy - g t Vy =0 t = Voy /g

this is the time to rise as the acceleration of gravity is the only serving time down is the same

tv = 2 t

tv= 2Voy /g

We calculate the horizontal distance

X = Vox t 3

Exercise gives the following condition

t = 5.5 Ymax

substitute and calculated in Equation 3

X = Vox t

5.5 Ymax = Vox tv

5.5 ( Voy2/ 2g) = Vox ( 2Voy /g) 5.5 Voy/2 = 2 Vox

5.5/4 VoSin = Vo Cos

1.375 Sin = Cos

Sin /Cos = 1/ 1.375

Tg = 0.7272

= 36°

6)

There are two parts to speed the jump (VJ) and swimming speed (Vs), we write these speeds on a Cartesian coordinate system

Vj = 6.47 m/s = 46.9°

Vs = 3.76 i m/s

We transform speed Vj to Vx e Vy

cos = Vjx/Vj Vjx = Vj Cos Vjx = 6.47 cos 46.9

Vjx= 4.42 m/s

Sin = Vjy/Vj Vjy = Vj Sin Vjy = 4.47 Sin 46.9

Vjy = 4.72 m/s

.a) Taking all the velocities in Cartesian system, we add separately in each axis

axis X

Vpx = (Vjx +Vsx)/2 Vpx = ( 4.42 +3.76)/2 Vpx = 4.09 ms

axisY

Vpy = ( Vjy +Vsy)/2 Vpy =( 4.72 +0 ) Vpy = 4.72 m/s

To find the answer we use the Pythagorean theorem

Vp2= Vpx2+ Vpy2 Vp2= 4.092+ 4.722 Vp2= 39

Vp = 6.246 m/s

Tg = Vpy/Vpx Tg = 4.72/4.09 tg = 1.154 = 49

.c)

We calculate the time to swim the distance L

Vs = X/t1 t1= L/Vs

We calculate the time to jump away L

X = Vjx t2 t2 = L/ Vj Cos

the total time

tT= t1 +t2 = L/Vs + L/ Vj Cos tT= L (1/Vs + 1/ Vj Cos )

tT= L ( 1/ 3,76 + 1/ 6.47 cos 46.9)

tT = = L 0.4921

We calculate the tempo if only he had swum 2L

Vs = x/t3 t3= 2L/ Vs

t3 = L 2 /3,76

t3 = L 0.5319

% = (tT/ t3 ) 100

% =( L 0.4921/ L 0.5319) 100 % = 0.4921/0.5319) 100

% = 92.5 %

Therefore the reduction ratio is

100- 92.5 = 7.5

The time was reduced by 7.5%

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.