Physics help!! test tomorrow trying to understand these types of problems?

Question

An airplane with a speed of 33 m/s is climbing upward at an angle of 44° counterclockwise from the positive x axis. When the plane's altitude is 740 m the pilot releases a package.
(a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.
m

(b) Determine the angle of the velocity vector of the package just before impact.
° clockwise from the positive x axis

How do I solve for the horizontal distance when there is only x-for distance in the kinematics equations........I have 740m and x for distances!! Please show your work so that I can figure out how to do these problems....thanks!

Answer 1

I initially had trouble understanding the way the question was phrased. Regardless, you do want to split up the 33 m/s into horizontal and vertical components.

Draw a coordinate plane. From the origin, draw a line and label it 33 m/s. Label the angle from the positive x axis as 44°. Draw a line straight down from the end of the 33 m/s line. You now have a right triangle with a hypotenuse of 33.

To find the vertical component:

sin 44° = y/33,

Vy = 22.92 m/s, upward.

To find the horizontal component:

cos 44° = x/33

Vx = 23.74 m/s, to the right.

When the package is released, it's going to be moving upward at 22.92 m/s. It wil decelerate to 0 m/s, then begin accelerating earthward. So, we have to split this up into two parts: the package moving up until its vertical velocity is 0; then the package moving down until it hits the ground.

For the first part, we have the initial velocity (22.92 m/s), we have the acceleration due to gravity (-9.8 m/s^2), and we have the final velocity (0 m/s). You will need to find out how much time before the vertical velocity is 0, then you need to find out how high the package traveled before it stopped (vertically).

We'll use the equation v = at, where v is your velocity, a is your acceleration and t is time. Let's suppose we started with an initial velocity of 0, and we wanted to know how long it would take to accelerate to 22.92 m/s.

22.92 = 9.8*t, so t1 = 2.34 s.

It would also stand to reason that if an object starts at 22.92 m/s and decelerated at the same rate, it would take the same amount of time.

Now we need to figure out how high it traveled:

y1 = (0.5)*a*t^2, where y1 is your distance, a is still 9.8 m/s^2, and t is the 2.34 s we just calculated. Solving for y1 gives us

y1 = 26.83 m.

Now, the package begins its descent toward earth. Again, we need to calcuate how long it takes to get there, and what its final vertical velocity is when it hits the ground.

It now has to fall 740 m + 26.83 m = 766.83 m.

We know the distance (766.83 m), we know the initial velocity (0 m/s), so we need to find how long it takes to reach the ground. Modifying the above equation:

766.83 = (0.5)*9.8*t^2,

so t^2 = 156.50, and

t2 = 12.51 seconds (not t^2: t2 signifies the downward time).

Let's find the vertical velocity of the package just before it hits the ground.

v = a*t

v = (9.8 m/s^2)*(12.51s) = 122.60 m/s, downward.

That's the vertical component.

Now, let's go all the way back to the beginning of our solution.

Vx = 23.74 m/s, to the right.

For the whole time that the package was airborne, it continued to travel at 23.74 m/s to the right. So to find the distance ...

d = r * t, or in this case d = r *(t1+t2)

d = 23.74*(2.34 + 12.51)

d = 352.54 m ... this is the answer to (a)
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(a) 352.54 m
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Now to determine (b). We know the vertical and horizontal velocities just before the package hits the ground:

Vertical velocity: 122.60 m/s, downward.

Horizontal velocity: 23.74 m/s, to the right.

Draw a vertical line, and label it 122.60 m/s. Draw a horizontal line from the bottom of that vertical line, and label it 23.74. Join the ends of the two lines with a third line. You now have another right triangle.

To find the angle, clockwise from the positive x axis, you can use the tangent function, since you know the two legs of the right triangle.

The tangent of the angle along the ground is opposite/adjacent, so

tan Ө = 122.60/23.74

inverse tan Ө = 5.16
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(b) Ө = 79.04 degrees.
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To find the magnitude of the velocity, you need to solve for the hypotenuse of the right triangle (I didn't see this specifically asked in the question, but here it is, anyway):

h^2 = (122.60)^2 + (23.74)^2.

h = 124.88 m/s
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I strongly encourage you to double-check my formulas and my math. It's been a while since I took physics.

I hope this was helpful. Good luck tomorrow.

Answer 2

What you need is to replace the negative thoughts with positive ones. That way you will be less nervous. If time is a problem, keep practicing and skip the harder ones (go back at the end). Another thing that might help is being prepared. If you're not prepared, it will make you nervous. So, study at least 2 days before the test, so you're not study the full 6 hours the day before the test. They say that after each science-related lecture, you should study 4 hours that night. The material will sink in and you will better understand the concepts for the test. Do all homework and study the material. Look over past quizzes and assignments and you should be ready to go!

Answer 3

Starting with the plane:

Vy0=(33m/s)sin(44deg)
Vx0=(33m/s)cos(44deg)

From 740m, we need to find out how long it takes the package to hit the ground. If we define up as positive y, then the package goes -740m along y. The initial velocity is above, the acceleration is due to gravity, g.

-740m = Vy0*(t) - (g/2)*t^2 solve for t, you want the positive one.

Using this t, the package goes Dx = Vx0*t horizontally (no acceleration)

Using this t you can find Vy = Vy0 - g*t. Vx = Vx0 and you can find the angle using trig.