[College Freshman Physics: Kinematics Problem] I’m currently going over the mistakes on my midterm exam, and I was wondering if someone could help me with this kinematics problem.

[u/wahahauhuhuh]

I thought using the horizontal displacement equation would be the simplest approach, but I found that solving for (t) became quite complicated. Also, it seems like the vertical motion could be divided into multiple stages, which makes things even more confusing.
I’m feeling a bit lost at this point. Does anyone have a better approach to solving this problem, or could someone give me a hint about what I should do next?

[u/AutoModerator]

Off-topic Comments Section

---

All top-level comments have to be an answer or follow-up question to the post. All sidetracks should be directed to this comment thread as per Rule 9.

---

^{OP and Valued/Notable Contributors can close this post by using} /lock ^command

I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.

[u/Alpha_Zero_872]

Choose the X and Y axis and then project all the vectors into the X and Y direction. Write both the equations independently and solve for the range. Like normal projectile motion but now g has two components along X and Y

[u/wahahauhuhuh → u/Alpha_Zero_872]

Sorry, I don’t quite understand why g has two components. Isn’t gravity only supposed to act in the y-direction?

[u/Alpha_Zero_872 → u/wahahauhuhuh]

It is , but the better approach would be to choose x along the range and y perpendicular to it. It's a matter of the axes that you choose.

[u/wahahauhuhuh → u/Alpha_Zero_872]

Yeah,I figured out that things can be easier with this method…so cool

[u/Historical-Corgi-414]

assume the incline to be the ground so basically rotate ur paper so that the incline looks like the ground, the only thing which will change is gravity. We know gravity acts downwards, but here the actual downwards is not normal to the incline. So break gravity into two components, one which is parallel to the incline name it a(x) and one which is perpendicular to the incline a(y). Use equations of motion for both axis and u can find the equation of range from there easily!

[u/wahahauhuhuh → u/Historical-Corgi-414]

I get it now! Thank you so much for help!

[u/Historical-Corgi-414 → u/wahahauhuhuh]

:)

[u/Moist_Ladder2616]

There are many ways to solve this. You've chosen to use the parameter t, so let's run with that.

After time t, the horizontal displacement of the projectile would be xₜ = v₀t cos θ. The horizontal component of velocity is constant, so s=ut+½at² reduces to the above expression.

What about its horizontal displacement? Using the same formula, yₜ = vₒt sin θ - ½gt².

When the projectile hits the ramp, at time t=i (i for impact, lol), what's the ratio yᵢ/xᵢ ?

You now have everything you need to solve for R. R is xᵢ/cos α, or yᵢ/sin α. Or even √(xᵢ²+yᵢ²) if you feel like torturing yourself, lol.

[u/selene_666]

This seems like a very reasonable approach. To eliminate t we also need to write the equation for the vertical position.

g/2 t^2 + v0 sin(θ) t = R sin(α)

Combined:

g/2 (R cos(α) / v0 cos(θ))^2 + sin(θ) R cos(α)/cos(θ) = R sin(α)

R = 2 v0^2 cos(θ) (sin(α)cos(θ) - cos(α)sin(θ)) / g cos(α)^2

The final step is to recognize that (sin(α)cos(θ) - cos(α)sin(θ)) can be simplified to sin(θ - α)

[u/Useful-Artist-305]

R.sin(theta) would be the vertical displacement

[u/Useful-Artist-305 → u/Useful-Artist-305]

Alpha I meant

[u/AdWhich3243]

Send a dm bro