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

Find the speed (2d)

  • Thread starter endeavor
  • Start date
176
0
1. Homework Statement
A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp (Fig. 3-46). (a) With what minimum speed must he drive off the horizontal ramp? The vertical height of the ramp is 1.5 m above the cars and the horizontal distance he must clear is 20m. (b) What is the new minimum speed if the ramp is now titled upward, so that "takeoff angle" is 10 degrees above the horizontal, and nothing else is changed?

http://img181.imageshack.us/img181/6396/chp3pro46sk5.th.png [Broken]

I need help on part b.
y0 = 1.5m
x = 20m.
Z = 10 degrees
2. Homework Equations
x = (v0 cos Z) t
0 = y0 + (v0 sin Z)t - 0.5*g*t2


3. The Attempt at a Solution
I have two equations in two variables, t and v0. So I solved for v0:
[tex]v_0^2 = (\frac{x^2}{2g\cos^2 Z})(\frac{1}{y_0 + x \tan Z})[/tex]

So, plugging in, I got 2.05 m/s for v0. But that has to be wrong, especially since my answer in (a) is 36m/s.

What am I doing wrong?
 
Last edited by a moderator:

Answers and Replies

176
0
I know what I did wrong. I put g in the denominator, but it should be in the numerator....
 
algebra mistake.... the g should be placed on the numerater...
 
1
0
How did you develop the equation to solve for the initial velocity?
 

Related Threads for: Find the speed (2d)

  • Last Post
Replies
1
Views
1K
Replies
7
Views
675
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
5K
Replies
2
Views
958
Replies
5
Views
2K
  • Last Post
Replies
5
Views
6K
Top