Projectiles dealing with a bullet

In summary, for a shell fired from the ground with an initial speed of 1.60*10^3m/s at an initial angle of 56° to the horizontal, the horizontal range can be found by resolving the horizontal component using cosine and the initial velocity, and then using the equation s=d/t to find the distance traveled. The amount of time the shell is in motion can be determined by using the kinematic equation v=u+at.
  • #1
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1. Homework Statement
A shell is fired from the ground with an initial speed of 1.60*10^3m/s at an initial angle of 56° to the horizontal.
(a) Neglecting air resistance, find the shell's horizontal range.

(b) Find the amount of time the shell is in motion.



Homework Equations



vector triangle sin and cosine

The Attempt at a Solution


all of my answers have been wrong i did the triangle and got hyp 1600 opp side 1326.46 bottom 894.7 from there i don't know where to go please HELP
 
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  • #2
You'll need to use the kinematic equations for projectile motion (horizontal and vertical components).
 
  • #3
i know that i just can't figure out the horiz and vertical components is
vertical
Vi:1600
thats all i know
 
  • #4
Would be better to:

1) Resolve the horizontal by doing Cos (theta) x initial velocity
2) v=u+at to get time in the air
3) Get horizontal component with trig like in 1)
4) distance bullet traveled is s = d / t rearranged to give d = s x t

That's all I can give you without giving you the answer on a plate :L
 
  • #5


I understand the importance of accurately solving problems and using the correct equations and methods. In this case, it seems that you may have made a mistake in your calculations. Let's take a closer look at the problem and see if we can figure out where you went wrong.

First, we need to understand the given information. We know that the shell is fired with an initial speed of 1.60*10^3m/s at an initial angle of 56° to the horizontal. This means that the initial velocity of the shell can be broken down into two components: a horizontal component and a vertical component. The horizontal component can be found using the cosine function, and the vertical component can be found using the sine function.

(a) To find the shell's horizontal range, we need to find the distance the shell travels in the horizontal direction before hitting the ground. This can be found using the equation: R = v0*cos(theta)*t, where R is the horizontal range, v0 is the initial velocity, theta is the initial angle, and t is the time the shell is in motion.

Plugging in the given values, we get R = (1.60*10^3m/s)*cos(56°)*t. Since we are neglecting air resistance, we can assume that the time the shell is in motion is the same as the time it takes for the shell to reach the ground. Therefore, we can use the equation h = v0*sin(theta)*t - (1/2)gt^2, where h is the height of the shell, g is the acceleration due to gravity (9.8m/s^2), and t is the time the shell is in motion.

Solving for t, we get t = (2*h)/(v0*sin(theta)). Plugging this into our first equation, we get R = (1.60*10^3m/s)*cos(56°)*(2*h)/(v0*sin(theta)). Simplifying this, we get R = (1.60*10^3m/s)*cos(56°)*2*h*sin(theta)/v0. Since sin(theta)/v0 is equal to the time of flight (t), we can rewrite this as R = (1.60*10^3m/s)*cos(56°)*2*h*t. Plugging in the given value for h (0m), we get R = 0m
 

1. How does the shape of a bullet affect its trajectory?

The shape of a bullet plays a significant role in its trajectory. A pointed bullet, for example, will have less air resistance and will travel farther and faster than a blunt bullet. This is because the pointed shape creates a more aerodynamic profile, reducing drag and allowing the bullet to maintain its velocity.

2. What factors affect the distance a bullet can travel?

The distance a bullet can travel is affected by several factors, including the initial velocity of the bullet, the angle at which it is fired, the air resistance, and the force of gravity. These factors can vary depending on the type of firearm and ammunition used, as well as environmental conditions.

3. How does the weight of a bullet impact its trajectory?

The weight of a bullet can greatly affect its trajectory. Heavier bullets will have more inertia and will be less affected by air resistance, allowing them to travel farther. However, lighter bullets may have a flatter trajectory and be more accurate at shorter distances. The weight of a bullet should be carefully considered when choosing ammunition for a specific firearm.

4. What is the relationship between bullet speed and accuracy?

The speed of a bullet is directly related to its accuracy. A faster bullet will have less time to be affected by external factors, such as wind or gravity, and will therefore maintain its trajectory better. However, too much speed can also cause a bullet to become unstable and less accurate. Finding the optimal speed for a specific firearm and bullet is crucial for achieving accuracy.

5. Can a bullet's trajectory be affected by external factors?

Yes, a bullet's trajectory can be greatly affected by external factors such as wind, air density, and gravity. These factors can cause a bullet to deviate from its intended path and can make it difficult to accurately hit a target. It is important for shooters to take these factors into account and adjust their aim accordingly in order to compensate for these external influences.

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