# Homework Help: Help With Projectiles Problem

1. Apr 14, 2008

### technology

[SOLVED] Help With Projectiles Problem

1. The problem statement, all variables and given/known data
A car rolls down a slope at 30° to the horizontal, and reaches 40 m/s just before plunging off a cliff 180m high. What is its velocity just before hitting the bottom?

2. Relevant equations
$$U_H = vcos\theta$$

$$U_V = vsin\theta$$

3. The attempt at a solution
I began the problem by finding intial velocity components:

$$U_H = 40cos30^o$$
$$U_H = 34.6 m/s$$
$$U_V = 40sin30^o$$
$$U_V = 20 m/s$$

I then found the time the projectile takes to fall:

$$s=ut+\frac{1}{2}at^2$$
$$-180=-20t-4.9t^2$$
$$4.9t^2+20t-180=0$$
$$t = 4.35s \mbox{(Using Quadratic Formula)}$$

The horizontal component stays constant at 34.6 m/s

I then calculated the final vertical component

$$V_v = U_v + at$$
$$V_v = -20 + (-9.8\times4.35)$$
$$V_v = 62.6 m/s \mbox {down}$$

I then constructed a vector diagram to calculate the resultant velocity. My working for the resultant velocity was as follows:

$$V_R = \sqrt{62.6^2 + 34.6^2}$$
$$V_R = 71.5 m/s$$

$$tan \theta = \frac{62.6}{34.6}$$
$$\theta = 61.1^o$$

Hence, according to my working, the velocity just before hitting the bottom is 71.5 m/s at 61.1° below the horizontal. However, the answer given on the answer sheet to these questions is 65.7 m/s at 72.3° below the horizontal. I am very stuck because I am not sure how to get this answer. Can anyone see where I have gone wrong? Any help would be much appreciated. Thanks very much in advance.

2. Apr 14, 2008

### Hootenanny

Staff Emeritus
You calculated the time incorrectly using the quadratic formula. However, there is no need to calculate the time of flight, instead, simply use a kinematic equation without time as a variable.

3. Apr 14, 2008

### RyanSchw

Ok, I’m still learning this stuff too and apparently I don’t know it as well as I thought. Given this problem I would have solved it the same way, and came up with the same answer as technology.

The only equation I can find that that relates velocity and time (it’s not in the book that I have) is

$$v_y^2 = v_{yo}^2 - 2gy$$

$$v_y = \sqrt{(-20 m/s)^2 - 2(9.8 m/s^2)(-180)} \approx 62.7~m/s$$

$$t = \frac{v_{yo} - v_y}{g} = \frac{(-20 m/s) - (62.7 m/s)}{9.8 m/s^2} \approx 4.35 sec$$
the same as the quadratic formula. =(

4. Apr 14, 2008

### Nabeshin

I can see nothing wrong with what either of you did because I am repeatedly getting the same result. Used work-energy and kinematics, both obtained Vf to be 71.6. The answer key is probably incorrect.

5. Apr 14, 2008

### Hootenanny

Staff Emeritus
Actually, scratch my previous post. You have calculated the time correctly, it was me that made the mistake (I has a sign wrong ). Your final answer and solution are correct.

Last edited: Apr 14, 2008
6. Apr 14, 2008

### robertm

Actually my solution matches the answer key technology,
heres how:

Vy,f^2 = Vi^2(sin0)-2g(change in y)

Vy,f^2 = 40^2(sin30)- 2(9.83)(180)

Vy,f^2 = 1600(.5)+3538.8

Vy,f^2 = 4338.8

Vy,f = 65.7

Don't second guess yourself hootenanny!

7. Apr 14, 2008

### Hootenanny

Staff Emeritus
Your solution is incomplete, what about the horizontal component of the velocity?

8. Apr 14, 2008

### kamerling

you should have

Vy,f^2 = 40^2(sin30)^2 - 2(9.83)(180)

that will get you the 62.7 m/s for Vy,f that RyanSchw and Technology also got.

9. Apr 14, 2008

### robertm

Oh crap i see i left off the square on the (sin0)^2. Sorry hootenanny

10. Apr 15, 2008

### technology

Thanks very much for the help guys, I think the answer key must be wrong. I appreciate the help.