Kinematics Problem (Find Speed)

AI Thread Summary
To determine the minimum speed required for a motorcyclist to clear a 48m gorge with a 15-degree launch angle and a vertical drop of 5.9m, kinematic equations and vector components must be applied. The initial calculations using the Pythagorean theorem yielded a distance of 48.36m, and time was calculated as approximately 3.14 seconds, resulting in a speed of 15.4 m/s. However, the importance of considering the launch angle and resolving the motion into horizontal and vertical components was emphasized, as this affects the overall trajectory. Participants in the discussion suggested that understanding projectile motion and the relevant SUVAT equations is crucial for solving such problems accurately. Ultimately, the correct minimum speed was identified as 25.4 m/s, highlighting the need for careful analysis of the problem's parameters.
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Homework Statement


A daredevil motorcyclist attempts to leap a 48m wide gorge. At the side where the cyclist starts, the ground slopes upward at an angle of 15 degrees. Beyond the far rim, the ground is level and is 5.9m below the near rim. what is the minimum speed necessary for the cyclist to clear the gorge?

Homework Equations



Kinematics Equations

s=d/t
d= vf(t)-0.5(a)(t)2

The Attempt at a Solution



First I just used Pythagorean therm to find Δd. 482+5.92= c2

c= 48.36

Used equation to find time

Δd = VfΔt-0.5(a)Δt2
48.36= -0.5(-9.8)Δt2
48.36 = 4.9Δt2
48.36/4.9=Δt2
9.86 =Δt2
+ - 3.14 = Δt

s=d/t
s=48.36/3.14
s=15.4 m/s

I think I was suppose to do something with 15 degrees.

I got the answer as 15 m/s, but I believe its wrong.
 
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Maybe you show us your workings and if we can spot any errors you might have made.
 
azizlwl said:
Maybe you show us your workings and if we can spot any errors you might have made.
First I just used Pythagorean therm to find Δd. 482+5.92= c2

c= 48.36

Used equation to find time

Δd = VfΔt-0.5(a)Δt2
48.36= -0.5(-9.8)Δt2
48.36 = 4.9Δt2
48.36/4.9=Δt2
9.86 =Δt2
+ - 3.14 = Δt

s=d/t
s=48.36/3.14
s=15.4 m/s

I think I was suppose to do something with 15 degrees.
 
1. Why vf=0?
2. The vertical height difference 5.9m below the launching point.
 
azizlwl said:
1. Why vf=0?
2. The vertical height difference 5.9m below the launching point.

How do I solve it without putting Vf as 0, and yes the vertical height difference is 5.9m below the launching point.
 
Have you learned about vectors?
\vec {A}=\vec {B}+\vec {C}

The motorcyle launches at 15° above the ground.
Since it is vector in 15° direction, you can resolved it to 2 components.
Y direction which affected by gravity and a constant X direction.
 
azizlwl said:
Have you learned about vectors?
\vec {A}=\vec {B}+\vec {C}

The motorcyle launches at 15° above the ground.
Since it is vector in 15° direction, you can resolved it to 2 components.
Y direction which affected by gravity and a constant X direction.

Can you please show me, how you would solve this please ??
 
If you do it once in lifetime, it is useful in giving you the answer.
I'll only tell you what prequisite knowledge that you should acquire in order to solve this type of question.

If you want to do physics problems, you should know about vectors and its operations.
Secondly you should know/remember/derive(this makes you less equations to remember) SUVAT equations.
 
This is a projectile motion problem. What can you tell us about projectile motion? What equations do you have that apply to projectile motion?
 
  • #10
If you have a question why we need vector components, then you are on the way to solve the problem.
http://imageshack.us/a/img805/2684/20946647.jpg
 
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  • #11
I'm really really sorry to hijack, but just wondering, is the answer 21.75m/s? Attempted it as practice, and just want to check the answer if I made any errors. Sorry! And Thanks!
 
  • #12
Nope. I get v0=25.4 m/s.
 
  • #13
Okay thanks lots! Time to redo then... :/
 
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