Minimum Speed for Stunt Car Jump over Parked Cars

[ciubba]

Homework Statement

A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp (Fig. 3–41). (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 22 m. (b) If the ramp is now tilted upward, so that “takeoff angle” is 7.0° above the horizontal, what is the new minimum speed?

Homework Equations

$$y=y_0+v_0t+.5at^2$$
$$v^2=v_0^2+2a(y-y_0)$$
$$x=v_0t$$

The Attempt at a Solution

The answer to part a is obviously 40m/s, but the answer to b eludes me.

I need v, so I defined v_x in terms of v via the kinematic equation $$x=v_0t$$

and then solved it for t

$$22=tcos\theta$$

$$t=\frac{22}{vcos\theta}$$

However, when I attempt to do this with v_y, I end up with a nasty quadratic

$$-4.9t^2+tvsin\theta+1.5=0$$

and solving this for t does not seem to be advisable.

 

[SammyS]

Homework Statement

A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp (Fig. 3–41). (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 22 m. (b) If the ramp is now tilted upward, so that “takeoff angle” is 7.0° above the horizontal, what is the new minimum speed?

Homework Equations

$$y=y_0+v_0t+.5at^2$$
$$v^2=v_0^2+2a(y-y_0)$$
$$x=v_0t$$

The Attempt at a Solution

The answer to part a is obviously 40m/s, but the answer to b eludes me.

I need v, so I defined v_x in terms of v via the kinematic equation $$x=v_0t$$ and then solved it for t
$$22=t\cos\theta$$
$$t=\frac{22}{v\cos\theta}$$
However, when I attempt to do this with v_y, I end up with a nasty quadratic
$$-4.9t^2+tv\sin\theta+1.5=0$$
and solving this for t does not seem to be advisable.

Hello, ciubba. Welcome to PF !

Why do you consider that quadratic equation to be nasty ?

You know that θ = 7° . Right ?

You should probably solve the other equation for t, in terms of v, then substitute that back into the quadratic, so there's only one variable.

 

[ciubba]

You should probably solve the other equation for t, in terms of v, then substitute that back into the quadratic, so there's only one variable.

I can't believe that never occurred to me-- I was going to solve them both for t, then set them equal to each other.

The answer is 23.88, which rounds to 24m/s with significant digits. Thanks!