• Support PF! Buy your school textbooks, materials and every day products Here!

Kinematics Problem (Find Speed)

  • Thread starter airking95
  • Start date
  • #1
4
0

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.
 
Last edited:

Answers and Replies

  • #2
1,065
10
Maybe you show us your workings and if we can spot any errors you might have made.
 
  • #3
4
0
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.
 
  • #4
1,065
10
1. Why vf=0?
2. The vertical height difference 5.9m below the launching point.
 
  • #5
4
0
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.
 
  • #6
1,065
10
Have you learned about vectors?
[itex]\vec {A}=\vec {B}+\vec {C}[/itex]

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.
 
  • #7
4
0
Have you learned about vectors?
[itex]\vec {A}=\vec {B}+\vec {C}[/itex]

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 ??
 
  • #8
1,065
10
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.
 
  • #9
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,538
1,149
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
1,065
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 [Broken]
 
Last edited by a moderator:
  • #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
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,538
1,149
Nope. I get v0=25.4 m/s.
 
  • #13
Okay thanks lots! Time to redo then... :/
 

Related Threads for: Kinematics Problem (Find Speed)

Replies
2
Views
3K
  • Last Post
Replies
9
Views
923
  • Last Post
Replies
1
Views
843
  • Last Post
Replies
3
Views
8K
  • Last Post
Replies
1
Views
1K
Replies
2
Views
3K
Top