What is the maximum altitude reached?

[Themachine8]

First Problem:

10 pts) A CESSNA Citation attains an airspeed of 680 km/hr. The plane sets out for a destination 1258 km to the north (a 0o course) but discovers that the plane must be headed 20o east of north to fly there directly. The plane arrives in 2.00 h. What is the wind speed?

What is the direction of the wind? Give your answer in degrees (o) where North = 0o, East = 90o, South = 180o, and West = 270o.

Second Problem:

(10 pts) A rocket is launched from rest and moves in a straight line at 55.0 o above the horizontal with an acceleration of 32.5 m/s2. After 36.0 s of powered flight, the engines shut off and the rocket follows a parabolic path back to earth. Find the time of flight from launch to impact.

What is the maximum altitude reached?

What is the horizontal distance between the launch pad and the impact point?

Thanks in advance for any sort of help...its due tomorrow at 8:30 am.

 

[Integral]

show us what you have tried.

The first problem is a simple application of Trig. Draw the triangle and do the arithmetic.

The second problem can be decomposed to 2 parts an x and a y problem. Use Trig to find the x and y components of acceleration. Then use the equations of a falling body to work out the rest.

 

[Themachine8]

My Work

Can you show me how to do the first problem...please, I am confused because of the vector addition.

For the second problem, here is my work, but when I enter it in online it says the answers are wrong (even after correcting for precision).

Start with a=d^2y/dt^2 and integrate twice. Taking 0 as the initial velocity and 0 as the initial position, we are left with (calling the distance "d"):

d = a*t^2

d = (32.5)(36)^2 = 42,100 m

But this has not been straight up, so the altitude is d*sin(55) = 34,000 m

The upward velocity is v = a*t*sin(55) = 958 m/s, and the upward acceleration is now g = -9.8 m/s^2

Integrating the same thing we did above twice and leaving in both constants for initial velocity and position, we have:

y = (1/2)a*t^2 + v_i*t + y_i

We need to find t when y = 0 (i.e. when the rocket hits the ground)

0 = (-4.9)*t^2 + (958)*t + 34,000

Solve however you like, finding t = 226 sec, to which we must add back in the initial 36 sec of powered flight, giving a total time of 262 sec.

To find the maximum altitude reached, reason that maximum altitude will occur when v = 0. Starting again with y_i = 34,000 m and v_i = 958 m/s, we use:

v = a*t + v_i

0 = (-9.8)*t + 958

Solving this we find the maximum altitude occurs at t = 98 sec after powered flight has ended.

Use the above formula for y, plugging in t = 98.

y = (-4.9)*t^2 + 958*t + 34,000 = 80,800 m is the maximum altitude attained.

Now, we need to find horizontal distance. Horizontally, when the rocket has d = 42,100 m traveled total:

x = d*cos(55) = 24,100 m

Using this as the initial position (x_i) and noting that the rocket (ideally, neglecting air resistance, etc.) travels at a constant velocity of v = 1170*cos(55) = 671 m/s in the x-dir for the rest of its flight, which is t = 226 sec long:

x = v*t + x_i = (671)(226) + 24,100 = 176,000 m

 

[Integral]

I did not check your computations on the 2nd problem, but your method is correct. One thing that is not clear to me. Does the initial acceleration of the rocket account for acceleration due to gravity? Try working it out by breaking the initial acceleration into components only subtract g from the y component. Because g is acting during the acceleration period also.


For the first problem, You can compute the total distance traveled (2hr*680km) and you have the direction traveled (20deg) . The wind velocity will be the difference between the path flown vector (1360km at 20deg) vs the path traveled vector ( 1258km at 0deg). Find the North and east components of the path flown vector then subtract from the path traveled. This is the components of the DISTANCE the wind pushed the plane. Assume the wind was constant the entire 2hr trip (divide distance by 2hr) to get the wind velocity vector.

Hope that is of some help.