Are these correct? need a little help

[barthayn]

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_f² = v_i² + 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_f² = (-.33.8)²+19.6*36
v_f² = 1143.08+705.6
v_f² = 1848.68
v_f² = 42.99m/s
v_f² = 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.

c² = 42.99² = (80cos25)²
c² = 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?

 

[anathema]

Hi there,

I am not able to view your vector diagram or the equations you used, so I cannot comment on those. However, I will provide some guidance and corrections for your solutions.

3a) The direction must be towards the southwest, not directly west. Also, the angle should be 9 degrees south of west, not 9 degrees south of southwest. So the final answer should be 9 degrees S of W.

3b) Your calculation for the time for the second leg of the trip is incorrect. The wind is blowing east, so it will actually decrease the time it takes for the plane to travel from Pittsburgh to Philadelphia. Your calculation should be t = d/v = 390/460 = 0.85 hours. So the total time for the flight would be 0.85 + 0.5 + 0.885 = 2.235 hours.

4a) Your calculation for the final velocity in the y-direction is incorrect. The y-component of the initial velocity should be positive, not negative. So it should be vf^2 = (33.8)^2 + 19.6*36 = 1848.68. The rest of your calculations are correct.

4b) Your calculation for the velocity is incorrect. You should use the final velocity in the x-direction, which is the same as the initial velocity. So the velocity of the shell as it strikes the ground is 80 m/s.

Overall, your solutions are mostly correct, but there are a few minor errors. I would suggest double-checking your calculations and making sure you understand the equations and concepts being used. Great job on completing most of your assignment!