# How Far Will a Steel Ball Travel at a 29-Degree Angle?

• am08
In summary, the ball would travel a distance of 5.108 meters if the same spring were aimed 29.0 degrees from the horizontal.
am08
[SOLVED] Projectile Problem

A small steel ball bearing with a mass of 16.0 g is on a short compressed spring. When aimed vertically and suddenly released, the spring sends the bearing to a height of 1.33 m. Calculate the horizontal distance the ball would travel if the same spring were aimed 29.0o from the horizontal.

Force has to be incorporated in here somehow so I was thinking:

F=mg --- (16*9.8) F=156.8N

Just use kinematics. The initial velocity is v0. What does vertical v0 have to be to take the ball to a height of 1.33m? Take the same v0 and solve the kinematic problem of vy0=v0*sin(29o) and vx0=v0*cos(29o). Hint: I say kinematics because the mass doesn't matter.

PaperAirplane
im having a little trouble with this...too many unknowns.

I know the maximum height: 1.33m and G = -9.8m/s^2

What equation do I use..

You have a few options to solve for the speed of the ball coming out of the cannon from the vertical case:

1.) Conservation of energy. All of the kinetic energy of the ball turns to potential energy at 1.33 m above ground. You can set up something that way.

2.) The kinematic equation vf^2 = vi^2 + 2ad. Plug in appropriate values and solve for vi, the initial velocity coming out of the cannon.

Then, you can use the range equation to solve for distance with your speed (if you don't know it, Range = (v^2*sin(2*theta))/g, where v is the muzzle velocity.

Hope this helps.

Last edited:
PaperAirplane
but vf and vi are unknown... so I can't solve for vi

Think about the vertical velocity of the ball at 1.33 m, the top of its upward path. It's moving from going upwards to downwards, right? So, at that instant, what is the velocity (or, what is vf, the final velocity)?

0 = vi^2 + 2(-9.81)*(1.33)

vi=5.108

Yup. Now use that velocity to solve for the range. You have the kinematics equations, but the easiest way is to just use the range equation.

Last edited:
PaperAirplane
terbum, thanks a lot man.

## 1. What is the "Steel Ball Projectile Problem"?

The "Steel Ball Projectile Problem" refers to the hypothetical situation in which a steel ball is launched from a certain height and angle with a given initial velocity, and the goal is to determine its trajectory and landing point.

## 2. How is the problem solved?

The problem is solved by using the principles of physics, specifically those related to projectile motion. This involves analyzing the forces acting on the steel ball, such as gravity and air resistance, and using equations to calculate its position and velocity at different points in time.

## 3. What factors affect the trajectory of the steel ball?

The trajectory of the steel ball is affected by several factors, including its initial velocity, launch angle, and the forces acting on it (such as gravity and air resistance). Other factors such as wind speed and direction can also have an impact.

## 4. What are some real-world applications of this problem?

The "Steel Ball Projectile Problem" has many real-world applications, such as in the design and testing of projectiles, such as missiles and rockets, as well as in sports such as baseball and golf. It is also used in engineering and construction, for example in determining the trajectory of falling objects.

## 5. Are there any limitations to this problem?

Some limitations of the "Steel Ball Projectile Problem" include assuming ideal conditions, such as a perfectly spherical steel ball, no air resistance, and a uniform gravitational field. In reality, these factors may vary and affect the accuracy of the solution. Additionally, other external factors that are difficult to account for, such as air turbulence, can also impact the trajectory of the steel ball.

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