# Help With Projectiles Problem

[SOLVED] Help With Projectiles Problem

1. Homework Statement
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. Homework 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.

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Hootenanny
Staff Emeritus
Gold Member
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.

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. =(

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.

Hootenanny
Staff Emeritus
Gold Member
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:
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! Hootenanny
Staff Emeritus
Gold Member
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! Your solution is incomplete, what about the horizontal component of the velocity?

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)
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.

Oh crap i see i left off the square on the (sin0)^2. Sorry hootenanny Thanks very much for the help guys, I think the answer key must be wrong. I appreciate the help.