Finding the Range of a Bullet Fired from a Sliding Incline

[Toranc3]

Homework Statement

A college student holds a gun in a horizontal position while sliding down a 55.0° incline at a constant speed of 8.00 m/s. How far will the bullet travel with a velocity of 5m/s if the student fires the gun when it is 10.0 m above the ground?

Homework Equations

y=y0+ v0y*t+1/2*g*t^(2)

The Attempt at a Solution

If he is sliding down an incline his acceleration would be gsin(theta) right? What if the slider was sliding at a constant speed as in this case? Would his acceleration just be g?

Looking to find the time it takes for him to go from 10m to 0m in the vertical motion.
y=y0 + v0t + 1/2g*t^(2)

What would g be in this case? Thanks!

 

[SammyS]

Homework Statement

A college student holds a gun in a horizontal position while sliding down a 55.0° incline at a constant speed of 8.00 m/s. How far will the bullet travel with a velocity of 5m/s if the student fires the gun when it is 10.0 m above the ground?

Homework Equations

y=y0+ v0y*t+1/2*g*t^(2)

The Attempt at a Solution

If a skier is skiing down an incline it's acceleration would be gsin(theta) right? What if the skier was skiing at a constant speed as in this case? Would his acceleration just be g?

Looking to find the time it takes for him to go from 10m to 0m in the vertical motion.
y=y0 + v0t + 1/2g*t^(2)

What would g be in this case? Thanks!

What does the skier have to do with this question regarding the bullet?

Are you wondering what would be the case is the student was sliding down the ramp without friction?

Then, yes, student's acceleration would be $\displaystyle \ g\sin(55^{\circ})\ .$

However, the student slides with constant speed. Therefore his/her acceleration is zero.​

 

[Toranc3]

What does the skier have to do with this question regarding the bullet?

Are you wondering what would be the case is the student was sliding down the ramp without friction?

Then, yes, student's acceleration would be $\displaystyle \ g\sin(55^{\circ})\ .$

However, the student slides with constant speed. Therefore his/her acceleration is zero.​

Oops sorry. I fixed that thanks. So for the equation y=y0+v0y*t+1/2*a*t^(2) a just goes to zero?
Since he is sliding down at constant speed he has zero acceleration along the incline but wouldn't he have acceleration pointing straight down?

 

[SammyS]

Oops sorry. I fixed that thanks. So for the equation y=y0+v0y*t+1/2*a*t^(2) a just goes to zero?
Since he is sliding down at constant speed he has zero acceleration along the incline but wouldn't he have acceleration pointing straight down?

What do you mean by "it" goes to zero? What is "it"?

Well, the acceleration of the student is zero, if that's what you mean, but that doesn't have much to do with solving this problem, or using that kinematic equation. The bullet will have acceleration of g, after being fired.

 

[Toranc3]

What do you mean by "it" goes to zero? What is "it"?

Well, the acceleration of the student is zero, if that's what you mean, but that doesn't have much to do with solving this problem, or using that kinematic equation. The bullet will have acceleration of g, after being fired.

Hey I got it. Idk what I was thinking before but I got it now. Thanks for your help.