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**[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**

[tex]U_H = vcos\theta[/tex]

[tex]U_V = vsin\theta[/tex]

**3. The Attempt at a Solution**

I began the problem by finding intial velocity components:

[tex]U_H = 40cos30^o[/tex]

[tex]U_H = 34.6 m/s [/tex]

[tex]U_V = 40sin30^o[/tex]

[tex]U_V = 20 m/s [/tex]

I then found the time the projectile takes to fall:

[tex]s=ut+\frac{1}{2}at^2[/tex]

[tex]-180=-20t-4.9t^2[/tex]

[tex]4.9t^2+20t-180=0[/tex]

[tex]t = 4.35s \mbox{(Using Quadratic Formula)}[/tex]

The horizontal component stays constant at 34.6 m/s

I then calculated the final vertical component

[tex] V_v = U_v + at [/tex]

[tex] V_v = -20 + (-9.8\times4.35) [/tex]

[tex] V_v = 62.6 m/s \mbox {down} [/tex]

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

[tex] V_R = \sqrt{62.6^2 + 34.6^2} [/tex]

[tex] V_R = 71.5 m/s [/tex]

[tex] tan \theta = \frac{62.6}{34.6} [/tex]

[tex] \theta = 61.1^o [/tex]

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.