Projectile motion car acceleration problem

[olliepower]

Homework Statement

A car is parked near a cliff overlooking the ocean on an incline that makes an angle of 29.3 degrees with the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline and has a velocity of 9m/s when it reaches the edge of the cliff. The cliff is 15.7m above the ocean. The acceleration of gravity is 9.8m/s^2

Homework Equations

\theta = 29.3
Velocity = V = 9 m/s
Initial height = 15.7 m
V_x = V cos /theta
V_y = V sin/theta
r_y = yo +V_oyt + 1/2gt^2 //equation for y component of the motion
r_x = V_xt //equation of x component of the motion

The Attempt at a Solution

1. Find x component
7.848263 = 9 cos (29.3)

2. Find y component
4.404444 = 9 Sin ( 29.3)

3. Find When Y motion = zero

0 = 15.7 -4.404444t - 1/2(9.8)t^2
t = 1.424256 //The car hits the water at this time

4. See how far the car goes in the X direction during the same time interval
X = 11.178448

 

[kuruman]

The y-component of the velocity and the acceleration of gravity must have the same sign because they are both downward.

 

[Spinnor]

I got 11.0 meters. Probably rounding differences. I used,

0 = 15.7 - 4.4*t - 4.9*t^2

and solved for t.

 

[olliepower]

The y-component of the velocity and the acceleration of gravity must have the same sign because they are both downward.

I did the actual calculation with -9.8m/s^2. I made a typo in the post. Fixed it now.

 

[olliepower]

I got 11.0 meters. Probably rounding differences. I used,

0 = 15.7 - 4.4*t - 4.9*t^2

and solved for t.

I took it out to 6 decimal units because that is what my HW system requires. I am still getting the wrong answer every time. Am I missing something?

 

[olliepower]

I emailed my professor and he responded with " You are solving a quadratic equation to find the time the car takes to hit the water, which is a recipe for wrong arithmetic. You can find the time using a first-order equation, and this is by far the best way to do it... much less chance of making a mistake."

What equation will let me know when height = 0 when i have the velocity?

 

[kuruman]

I solved the quadratic and I got 10.957633 m. It's a ridiculous number of decimal places, but that's what it is. Probably your professor is referring to the following derivation for the time.

1. Start with v_y² = v_0y² + 2aΔy
2. Solve for v_y by taking the square root.
3. Substitute the result in v_y=v_0y + at
4. Solve the resulting equation for the time.

I am not convinced that this four-step derivation provides "much less chance for making a mistake" than the one-step solution of the quadratic. Take your pick.

 

[olliepower]

Thanks for the help guys. I finally got it. U guys are the best!