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

[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

Im now totally lost, please help.

 

[Dick]

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.

 

[am08]

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..

 

[terbum]

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.

 

[am08]

but vf and vi are unknown... so I can't solve for vi

 

[terbum]

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)?

 

[am08]

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

vi=5.108

 

[terbum]

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.

 

[am08]

terbum, thanks a lot man.

I really appreciate your help