Projectile motion, missed two weeks, fell behind, no book, Please help Thank you

[Bearman94]

Homework Statement

Hi guys, I am not one to ask for help but i missed ALOT of school because I got sick and I was wondering if anyone could solve three problems for me and help me out!
Thanks in advance !
Problem 1-
A plane is dropping a supply of coconuts to survivors of the USS Minnow, who are stranded on a desert island. If the plane is 2km in the air and is traveling at 200 m/s, How far in front of the island should the supplies be dropped?
Problem 2- A watermelon is thrown off the 10th (40 meters above the ground) floor of a building. It is launched with a speed of 30 m/s at an angle of 40 degrees above the horizontal. How far away does it land?
Problem 3- A football is kicked at 25 m/s at an angle of 50 Degrees.
How fast is the football traveling when it strikes the ground and what direction is it traveling.
show work please.
and once again, I'm very grateful to anyone who will help me out! I am trying to do better in this class and I just happened to get sick:/

Homework Equations

All can be found here - http://tutor4physics.com/projectilemotion.htm
And on other sites, but none seem to copy + Paste
Equation of Trajectory (Path of projectile)

The Attempt at a Solution

AND I have read the FAQ and the statement, however, I am really in a bad position here, I have attemped these but I know that it is not right, and Posting it would just probably make me look dumb and contribute nothing. I know how projectile motion works and I have done about 80% of my Homework, but these 3 problems are stumping me.

For the Watermelon Problem All i know is the X and Y Componants
Which are Cos(40Deg) * 30 M/s and Sin(40Deg) * 30 m/s
And that A=Change in V / T
so
- 9.8m/s^2= -30 /t
-9.8m/s^2 * t = -30
T = 3.06
And then I know you must use the 40m To make another vector to complete the problem.
For the Football Problem I know.
How Long it is in the air- 3.9 Seconds
Because Ttotal=38.3m/s / 9.8 m/s
How high it goes= 18.7m
How far away does it land 62.8 m
because Range= (25M/S)^2 * Sin2(50) /
9.8 M/s^2But for the plane problem I am totally lost

 

[Bearman94]

/lk;

 

[Chopin]

The key to all projectile motion problems is to remember that the x component of the velocity will remain unchanged from start to finish, and the y component of the velocity will decrease according to gravitational acceleration. Therefore, for any problem, you should first begin by determining the initial x and y velocities, and then you will have two independent problems to solve--one in x, and one in y.

For 1, the y velocity begins at 0. The plane is 2km up, so first of all, you should try to figure out how long it will take to hit the ground. Once that's done, try to determine what the horizontal velocity is, and then you should have enough information to determine how far forward the coconuts will go.

For 2, you've already determined the watermelon's initial vertical velocity. For a moment, just pretend the watermelon was thrown straight up into the air, instead of at an angle. Again, try to figure out how long it takes to go up into the air, fall back to its starting point, and then fall an additional 40m.

3 is exactly the same as 2, except the football only goes down to 0, not any further.

In all cases, time is the parameter that links the x and y equations together. Find that, and you should be able to compute all the other missing quantities off of it. The equations at http://physics.bu.edu/~redner/211-sp06/class01/equations.html might prove useful--they seem a bit more straightforwardly written than the ones on the page you linked. For each problem, write down each variable that you know, and determine which equation has only one unknown quantity left in it.

 

[azizlwl]

There are 2 component of the flight of the coconuts.
1. Horizontal component, no acceleration
2. Vertical component with acceleration due to gravity.

V=ut+1/2(at²)
2000=1/2(10 t²) taking a=10m/s² and they dropped, with no initial vertical velocity.
400=t² => t=20sec

H=ut+1/2(at²) where a=0
H=20(200)= 4km.

 

[asteriatic]

on how to even start it. Can someone please help me and show me the steps to solve these problems?

Dear student,

I understand that you have missed two weeks of school due to illness and are now feeling overwhelmed and behind in your studies. I can assure you that it is completely normal to feel this way when you have missed important lessons and materials. However, I am glad that you have reached out for help and I am here to assist you.

Firstly, let's address the problem with the plane dropping coconuts. This problem involves projectile motion, which is the motion of an object in the air under the influence of gravity. To solve this problem, we need to use the equations of projectile motion, which can be found on various websites including the one you have mentioned.

The first step is to identify the known variables in the problem. We know that the plane is 2km in the air and is traveling at 200 m/s. We also know that the survivors are on a desert island and we need to find the distance in front of the island where the supplies should be dropped. This distance can be represented as "x" in our problem.

Next, we can use the equation of trajectory to determine the distance x. This equation is as follows:

x = (V^2 * sin 2θ) / g

Where:
V = initial velocity (in this case, the velocity of the plane)
θ = angle of projection (in this case, the angle at which the supplies are dropped)
g = acceleration due to gravity (which is 9.8 m/s^2)

Substituting the known values in the equation, we get:

x = (200 m/s)^2 * sin 2θ / 9.8 m/s^2

Now, we need to find the angle at which the supplies are dropped. This can be done by using trigonometric ratios. Since we know the height of the plane (2km), we can use the tangent ratio to find the angle. The tangent ratio is given by:

tan θ = opposite/adjacent

In this case, the opposite side is 2km and the adjacent side is the distance x. Substituting these values, we get:

tan θ = 2km / x

Rearranging the equation, we get:

θ = tan^-1 (2km / x)

Now, we can substitute this value of θ