A 2D Kinematics Question and Relative Velocity Question

[baillik]

Question 1:
A catapult on a cliff launches a large round rock towards a ship on
the ocean below. The rock leaves the catapult from a height H of 32.0 m above sea level, directed at an angle above the horizontal with an unknown speed v0. The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 160 m. Assuming that air friction can be neglected,
A. calculate the value of the angle (in degrees).
B. Calculate the speed at which the rock is launched.

I've looked at similar questions posted on this forum, but I am still unsuccessful at getting the answer.

Relevant Equations:
X = Xo+ Vot+.5at^2

V=Vo+at

V^2= Vo^2 + 2a (X-Xo)

Okay, here is what I have done:
A. x = xo + vixt + 1/2at^2
160 = vix(6.00) + (1/2)(-9.81)(6.00)^2
vix=56.1m/s
y=yo + viyt + 1/2at^2
-32 = viy(6.00) + (1/2)(-9.81)(6.00)^2
viy=24.1
theta = 24.1/56.1tan^-1 = 23.2 deg
but the answer says it should be 42.1 deg, i don't know where i am making my mistake?

B. (viy)^2 + (vix)^2 = (vi)^2
SQRT((56.1)^2 + (24.1)^2) = 61.0m/s
but the answer should be 35.9m/s, again i can't find my mistake.


Question 2:
A wind is blowing directly from east to west. The pilot of a small plane finds that if he points the nose of the plane 28.4 ° north of east, his velocity with respect to the ground is in the direction 56.7 ° north of east. The speed of the plane with respect to the air is 129 m/s.
A. Taking North to be the y-direction and East to be the x-direction, what is the y-component of plane's velocity with respect to the ground?
B. What is the magnitude of the plane's velocity with respect to the ground?

Relevant Equations:
I don't know, haha, no idea where to go with this problem

What I did:
I really don't know what to do with this one, I tried drawing a diagram but that didn't really help me out.

Any help would really be appreciated!

 

[zgozvrm]

There is no acceleration in the x direction (only in the y-direction). So your equation for [itex]V_{ix}[/tex] is incorrect.

 

[baillik]

There is no acceleration in the x direction (only in the y-direction). So your equation for [itex]V_{ix}[/tex] is incorrect.

oh, okay, i see that, thanks!

any idea for the second question?

 

[zgozvrm]

The trick here is in the wording of the question...
It says that the speed of the plane with respect to the air is 129 m/s.

Since there is a wind which blows the plane off course, that means that the air is moving.

Suppose there where 2 planes heading directly toward each other with no wind (plane A at 150 m/s and plane B at 200 m/s).
From the perspective of the ground, plane A would be traveling at 150 m/s.
(In other words, the speed of plane A with respect to the ground is 150 m/s).
The speed of plane B with respect to the ground is 200 m/s.
But the speed of plane A with respect to plane B would be 350 m/s.

That is, to someone sitting in plane B, plane A would appear to be coming at them at 350 m/s.

But in this case the plane's heading and the wind's direction are at an angle to each other (not directly in line). So the question now becomes, "What is meant by 'the speed of the plane with respect to the air'?"

Since 129 m/s [itex]\approx[/tex] 4.8 mph, we know that this is not the plane's velocity before taking the wind into account, since that is too slow for a plane to remain in flight.

Therefore, the 129 m/s speed must refer either to the plane's final velocity, or the x-component of it's final velocity. But part B is asking for the magnitude of the plane's final velocity, so ...


Either that, or the question is poorly (incorrectly?) worded.

 

[rwisz]

Hi there,

For Question 1, it seems like you have made a mistake in your calculation for the angle. The correct equation to use for finding the angle is tan^-1(viy/vix), not viy/vix. So, the correct calculation would be tan^-1(24.1/56.1) = 23.2 degrees.

For the speed, you have also made a mistake in your calculation. The correct equation to use is v = sqrt((vix)^2 + (viy)^2), not (viy)^2 + (vix)^2. So, the correct calculation would be v = sqrt((56.1)^2 + (24.1)^2) = 60.9 m/s.

For Question 2, you can use the equation v = u + w, where v is the velocity with respect to the ground, u is the velocity with respect to the air, and w is the velocity of the wind. Since the wind is blowing from east to west, the y-component of the wind's velocity is 0. Therefore, the y-component of the plane's velocity with respect to the ground would also be 0.

The magnitude of the plane's velocity with respect to the ground can be found using the Pythagorean theorem, as v = sqrt((v_x)^2 + (v_y)^2). So, the magnitude of the plane's velocity with respect to the ground would be 129 m/s, since the x-component of the plane's velocity with respect to the ground is equal to the speed of the plane with respect to the air.

I hope this helps! Remember to always double check your equations and units to avoid mistakes. Keep up the good work!