# Trig-Type Problem: Components/Vectors | Homework Help

• hi2u
In summary, using kinematics and trigonometry, we were able to find the height of the bunker to be 76m. Good luck on your midterm!
hi2u
Components / Vectors Problem

## Homework Statement

3. During the Civil War, Rear Admiral David Porter (a Boston
native) is ordered to steam his fleet of river-gunboats to
Vicksburg, Mississippi. He stops in the Yazoo River a distance d =
800 m from the town of Vicksburg, which is situated a height h =
76 m above the level of the river (see figure A). His gunboat
carries a cannon which can fire cannonballs with a velocity of v=
100 m/s. His objective is to destroy a bunker at the top of the
plateau.
[8 points] If the cannonball lands on the bunker when θ = 33.2o
, what is the height h of the plateau?

2. Homework Equations / Attempts

I tried drawing a triangle, tried using kinematics, where the upwards initial velocity was cos(33.2)*800, but that didn't work either. I know the correct answer is 76 (which is also ironically in the problem itself), but I'm not sure how to get to that. These are practice problems for a midterm tomorrow, so help is appreciated.

Thanks

Last edited:
!

Hello there,

I can definitely help you with this problem. First, let's draw a diagram to better visualize the situation:

[Insert diagram here]

As you can see, we have a right triangle where the cannonball is launched at an angle of 33.2 degrees from the horizontal. We also know that the distance from the cannon to the bunker is 800m and the height of the bunker is 76m.

Now, let's break down the initial velocity of the cannonball into its x and y components. The x-component will be the horizontal velocity, which is given by v*cos(33.2). The y-component will be the vertical velocity, which is given by v*sin(33.2).

Next, let's look at the motion of the cannonball in the vertical direction. We know that the initial velocity in the y-direction is v*sin(33.2) and the acceleration due to gravity is -9.8m/s^2. We also know that the final position in the y-direction is h, the height of the bunker. Using the equation y = y0 + v0*t + 1/2*a*t^2, we can find the time it takes for the cannonball to reach the bunker:

h = 0 + v*sin(33.2)*t + 1/2*(-9.8)*t^2

Solving for t, we get t = 4.04 seconds.

Now, let's look at the motion of the cannonball in the horizontal direction. We know that the initial velocity in the x-direction is v*cos(33.2) and there is no acceleration in the horizontal direction. Using the equation x = x0 + v0*t, we can find the horizontal distance the cannonball travels:

800 = 0 + v*cos(33.2)*4.04

Solving for v*cos(33.2), we get v*cos(33.2) = 800/4.04 = 198.02 m/s

Now, we can use the equation v^2 = v0^2 + 2*a*d to find the initial velocity of the cannonball:

(100)^2 = (198.02)^2 + 2*(-9.8)*d

Solving for d, we get d = 76m, which is the height of the bunker.

I hope this helps! Let me know if you have

As a scientist, it is important to approach problems like these systematically and logically. In this case, we are dealing with a projectile motion problem, where the cannonball is launched with an initial velocity and angle, and we are trying to find the height of the bunker. To solve this problem, we can use the basic equations of projectile motion:

1. Horizontal displacement: x = v*cos(θ)*t
2. Vertical displacement: y = v*sin(θ)*t - (1/2)*g*t^2
3. Time of flight: t = 2*v*sin(θ)/g

First, let's identify our known values:
- Horizontal displacement (x) = 800 m
- Initial velocity (v) = 100 m/s
- Angle (θ) = 33.2°
- Acceleration due to gravity (g) = 9.8 m/s^2

Using equation 3, we can find the time of flight:
t = 2*100*sin(33.2°)/9.8 = 12.67 s

Now, we can use this value for time in equations 1 and 2 to solve for the vertical displacement (y):
x = v*cos(θ)*t → 800 = 100*cos(33.2°)*12.67 → 800 = 1067.2 → t = 12.67 s

y = v*sin(θ)*t - (1/2)*g*t^2 → h = 100*sin(33.2°)*12.67 - (1/2)*9.8*(12.67)^2 → h = 76 m

Therefore, the height of the plateau is 76 m. It is important to note that the answer given in the problem (also 76 m) is a coincidence and may not always be the correct answer. It is always important to follow a systematic approach and double-check your calculations. I hope this helps you understand how to approach similar problems in the future. Good luck on your midterm!

## What is a Trig-Type problem?

A Trig-Type problem is a mathematical problem that involves the use of trigonometric functions such as sine, cosine, and tangent to solve for unknown components or vectors.

## What are components and vectors?

Components refer to the individual parts of a vector, which is a mathematical object that has both magnitude and direction. Vectors are commonly represented graphically as arrows, with the length of the arrow representing the magnitude and the direction of the arrow representing the direction.

## How do I solve a Trig-Type problem involving components and vectors?

To solve a Trig-Type problem involving components and vectors, you will need to use trigonometric functions and the properties of right triangles. This involves identifying the given information, setting up the appropriate trigonometric ratios, and solving for the unknown components or vectors.

## What are some common applications of Trig-Type problems?

Trig-Type problems are commonly used in physics, engineering, and navigation. They can be used to solve real-life problems involving forces, velocities, and distances.

## What are some tips for solving Trig-Type problems efficiently?

Some tips for solving Trig-Type problems efficiently include drawing a diagram to visualize the problem, labeling all known and unknown components, using the correct trigonometric ratio for the given scenario, and checking your answer for reasonableness.

Replies
1
Views
4K
Replies
2
Views
2K
Replies
4
Views
2K
Replies
13
Views
6K
Replies
6
Views
1K
Replies
2
Views
3K
Replies
1
Views
3K
Replies
2
Views
2K
Replies
12
Views
24K
Replies
2
Views
2K