A stunt driver wants to make his car jump over eight cars parked side by side below a horizontal ramp.

(a) With what minimum speed must he drive off the horizontal ramp? The vertical height of the ramp is 1.5m above the cars, and the horizontal distance he must clear is 20m.

(b) If the ramp is now tilted upward, so that "takeoff angle" is 10 degrees above the horizontal, what is the new minimum speed?

this is from that white book that i'm also using...

the odd answers in the book are in the back in which this problem is one I belive

You know the horizontal distance it must go.

Horizontal distance=vi*time
verticalpositionfinal=verticalpositioninitial -4.9 t^2
0=1.5-4.9 t^2 solve for t, then put that into the horizontal distance equation and solve for vi.

Now if you tilt it, then the vihoriontal is viCos10, and vivertical is not zero, but ViSin10.

Work it the same way, it is a little tricker on the algebra, but it works cleanly.

I don't know how the author of the Giancoli text reached 20 m/s as the answer to the second part of this questions. Nevertheless, I do know that the solution Bobpursley suggests here doesn't work. I find it interesting that there isn't one straightforward solution to this question on the entire internet. I've been working on the problem for two days, and none of the strategies I have tried give me 20 m/s.

To call the algebra on this question convuluted is a profound understatement. Nevertheless, I have solved it.

The range equals 20 meters and the initial velocity of the car along the x-axis is 0.98Vi (or the Vicos10). Since the combined time the car travels up and down on the y-axis equals the time the car travels the entire way along the x-axis, you can set the equation as follows:

time = time

or

20 / 0.98Vi = time on y-axis

Divide 20 by 0.98 to find that the ratio of the initial velocity is:

20.4 / Vi

...or more simply put, the initial velocity if 20 m/s.

To find the minimum speed the stunt driver must have to clear the cars, we can use the principles of projectile motion. Let's break down the problem into two parts.

(a) First, let's consider the motion in the horizontal direction. The horizontal distance the car must clear is 20m. We can use the equation:

horizontal distance = initial horizontal velocity × time

Since there is no horizontal acceleration, the initial horizontal velocity remains constant during the entire jump. Therefore, we can rewrite the equation as:

20m = initial horizontal velocity × time

Since the time is the same for both the horizontal and vertical motion, we can use this equation to find the horizontal velocity.

(b) Now, let's consider the motion in the vertical direction. The vertical height of the ramp is 1.5m. We can find the time it takes for the car to reach the maximum height using the equation:

vertical displacement = (initial vertical velocity × time) + (0.5 × acceleration × time²)

Since the initial vertical velocity is zero (the car starts from rest in the vertical direction), the equation simplifies to:

1.5m = 0.5 × acceleration × time²

Next, we need to find the acceleration. The only force acting on the car in the vertical direction is gravity, which causes an acceleration of 9.8 m/s² downward. Therefore, the equation becomes:

1.5m = 0.5 × (9.8 m/s²) × time²

We can rearrange this equation to solve for time:

time = √(2 × 1.5m / (9.8 m/s²))

Now that we know the time, we can substitute it into the equation we found earlier to determine the horizontal velocity:

20m = initial horizontal velocity × time

Substituting the value of time, we can solve for the initial horizontal velocity.

Now let's move on to part (b), where the ramp is tilted upward at an angle of 10 degrees.

The only change in this case is the takeoff angle of the ramp. To find the new minimum speed required, we can use the same principles of projectile motion.

Again, we can consider the horizontal motion and the vertical motion separately.

In the horizontal direction, the equation remains the same:

20m = initial horizontal velocity × time

The difference now is that the initial velocity has a vertical component. We can find this component using trigonometry:

vertical component of initial velocity = initial speed × sin(takeoff angle)

In this case, takeoff angle is 10 degrees. With this information, we have the vertical component of the initial velocity.

In the vertical direction, the equation is similar to before:

1.5m = 0.5 × acceleration × time²

The acceleration due to gravity remains the same, and the time can be found using the formula:

time = √(2 × 1.5m / (9.8 m/s²))

Now, we need to find the new minimum speed. This speed is the magnitude of the initial velocity, which can be found using the Pythagorean theorem:

initial speed = √((horizontal component of initial velocity)² + (vertical component of initial velocity)²)

Substituting the values we found earlier, we can calculate the new minimum speed.