Final Velocity with barely anything AP Phys help

[anshu]

A boy fires a "BB" gun at an angle of 28o above the horizontal. The "BB" strikes a target that is a vertical distance of 1.6 meters above and 12.8 meters to the right of the muzzle (exit point) of the gun. Assuming the bullet has a negative vertical velocity when it strikes the target, calculate the bullet's velocity when it leaves the gun.

i tried everything that i could think of, using trig to figure out

Vx=Vy/tan(28) and therefore Vy/tan(28)=12.8

anything that i would plug into the y=VoT + 1/2(a)(Tsqrd) i would get a negative square root


Thanks in Advance
Anshu

 

[maverick280857]

Welcome to PF Anshu...

Well the obvious line of attack for such a problem would be to write the expressions for the horizontal and vertical distances in terms of time and eliminate time to get a trajectory equation of the form,

$$y - y_{0} = (x-x_{0})\tan \alpha -\frac{1}{2}\frac{g(x-x_{0})^{2}}{V_{0}^2}\sec^2\alpha$$

and then plug in all that you have to solve for the initial velocity. (First, try and derive the above equation to convince yourself that it is true).

The other method is to solve for Vx and Vy somehow and relate them through the angle they make. My question here: What do you think the angle between Vx and Vy will be at the time the shot hits the wall? Will it be 28 degrees? Are you sure of that?

Also, you have mentioned an equation for y, the vertical distance. What do you propose to do with it?

Hope that helps...

Cheers
Vivek

 

[M98Ranger]

Hi Anshu,

First of all, great job in trying to use trigonometry and kinematic equations to solve this problem! It shows that you have a good understanding of the concepts.

To calculate the final velocity of the bullet, we can use the following kinematic equations:

1. Vertical displacement equation: Δy = voyt + 1/2gt^2
2. Horizontal displacement equation: Δx = vox t
3. Vertical velocity equation: voy = voy + gt

Using the given information, we can set up the following equations:

1. Δy = 1.6 m, voy = unknown, t = unknown, g = -9.8 m/s^2 (since the bullet has a negative vertical velocity)
2. Δx = 12.8 m, vox = unknown, t = unknown
3. voy = unknown, voy = unknown, g = -9.8 m/s^2

Now, we can use trigonometry to find the initial vertical and horizontal velocities of the bullet:
voy = vox tan(28)
vox = voy/cos(28)

Substituting these values into the equations above, we get:

1. 1.6 = (voy tan(28))t - 4.9t^2
2. 12.8 = (voy/cos(28))t

Solving for t in the second equation and substituting it into the first equation, we get:

1. 1.6 = (voy tan(28))(12.8/voycos(28)) - 4.9(12.8/voycos(28))^2
2. 1.6 = 12.8tan(28)/cos(28) - 4.9(12.8)^2/voycos^2(28)

Simplifying and solving for voy, we get:
voy = -17.8 m/s (negative sign indicates that the velocity is downwards)

Now, we can use this value of voy to find vox:
vox = voy/cos(28) = -17.8/cos(28) = -20.1 m/s

Therefore, the final velocity of the bullet when it leaves the gun is:
v = √(vox^2 + voy^2) = √((-20.1)^2 + (-17.8)^2) = 26.3 m/s

I