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**Are these right?**

Hi, I just did 2/3 of my assignment for physics and had trouble with the first question. I was wondering if I was correct. Please look over my work.

## Homework Statement

3. Two planes fly from Toronto to Philadelphia. Plane A flies via Pittsburgh where as passengers on plane B have a direct flight. Pittsburgh is 350 hm due south of Toronto and 390 km due west of Philadelphia. The airspeed of both planes is 400km/h and a steady wind is blowing from the east at 60km/h.

a)What direction must the pilot point the plane flying from Toronto to Pittsburgh. Include a vector diagram of velocities.

b) How long will the entire flight take for plane A assuming a 0.50h layover in Pittsburgh?

4. A shell is fired from a cliff that is 36m above the horizontal plane. The muzzle speed of the shell is 80m/s and it is fired at an elevation of 25

^{o}above the horizontal.

a) Determine the horizontal range of the shell.

b) Determine the velocity of the shell as it strikes the ground.

## Homework Equations

v = d/t

y = ((v

_{i}+v

_{f})/2)t

v

_{f}

^{2}= v

_{i}

^{2}+ 2ay

## The Attempt at a Solution

3. a) I drew a vector diagram. The hypothesis of the side of the triangle was 400km/h whilst the opposite was 60km/h. The other side (adjacent) turned out to be around 395km/h. This was a right angle triangle.

I used TANθ=o/a

TANθ = 60/395

TANθ = 0.1517

θ = 9

^{o}

Therefore the plane must travel 400km/h [9

^{o}S of W] To reach Pittsburgh. Is this correct?

b) I used the same diagram and got t = d/v

t = 350/395

t = 0.885h

V

_{total}= v

_{normal}-v

_{air}

V

_{total}= 400 - 60

V

_{total}= 340km/h

t= d/v

t= 390/340

t= 1.14

t

_{total}= 1.14+.5+0.885

t

_{total}= 2.5h

Therefore it would take 2.5h with the 0.5h layover.

4 a) let up be negative

v

_{f}

^{2}= (-.33.8)

^{2}+19.6*36

v

_{f}

^{2}= 1143.08+705.6

v

_{f}

^{2}= 1848.68

v

_{f}

^{2}= 42.99m/s

v

_{f}

^{2}= 43m/s

t = ((2y)/(v

_{i}+v

_{f}))

t = 72 / 76.8057

t = 0.93s

d = tv

d = 80cos25*0.93

d = 68m

Therefore the horizontal range of the shell is 68m.

I used my final velocity for the y-axis to get two sides of the diagram because horizontal does not change.

c

^{2}= 42.99

^{2}= (80cos25)

^{2}

c

^{2}= 1848.68

c = 42.99m/s

TANθ = o/a

TANθ = (80cos25)/42.99

TANθ =1.6863

θ = 59.33

θ = 59

Therefore the shell hit the ground at 84m/s [59

^{o}below the horizontal].

Are these all correct? Did I do 3 a correct? Where can I improve?

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