Question:

A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp (Fig. 3-41).

- 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 22 m.
- If the ramp is now tilted upward, so that "takeoff angle" is $7.0^\circ$ above the horizontal, what is the new minimum speed?

Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.

Quick Answer:

- $4.0 \times 10^1\textrm{ m/s}$
- $24 \textrm{ m/s}$

## Comments

why do you do these things?

Hi kmoons25, Well, I know physics can be confusing. When you have a more specific question, just let me know.

All the best,

Mr. Dychko

I originally tried using the Range formula which gave me a slightly wrong answer. I'm assuming it's not applicable in this situation because of the 1.5 m drop?

Hi idan, yes you're exactly right. The range formula was derived using the assumption that the final and initial heights are the same. Since that's not the case here, as you say, the range formula doesn't apply.

Cheers,

Mr. Dychko

For part b why didn't we also find out Vx. Why did we find Vy

Hi suriyak786, thank you for your question. At 3:20 we made use of $v_x$ to create an expression for $t$ in terms of things that we know, such as $x$ and $\theta$, and the one thing we don't know, $v$. That was then substituted into the vertical displacement formula, which only has $v_y$ in it since only the vertical component of velocity affects the vertical displacement of the car. Does that help?

All the best,

Mr. Dychko