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A projectile is launched over level ground at $35 \frac{m}{s}$ at an angle of $24^{\circ}$ above the horizontal. Friction is negligible. What is the velocity (magnitude and direction) of this projectile $2.5$ s after launch?

For the magnitude do they mean the overall velocity at $2.5$ seconds $(\cos(24)35)$ in or do they mean the magnitude of the displacement: $\cos(24)35 \cdot 2.5$ ?

For the direction part do you assemble a triangle of the the combined velocity and the x and y components. Then use the cos law to solve for cos of the angle? Is this completely wrong? The answer on my homework differs from mine, but maybe thats just cause my math skills are questionable. Can anyone help me out with this and solve it step for step with an answer at the end? Thanks soooooo much :)!!!!!!

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Remember to include the force exerted by gravity on the projectile. This will affect both the magnitude and angle of the velocity. –  user12998 Aug 27 '11 at 23:12
    
Hint: The horizontal component of velocity is $35\cos(24^\circ)$. For the vertical component, initially it is $35\sin(24^\circ)$ (if up is positive), but changes because of gravity. Once you find the two components $a$ and $b$, the magnitude of the velocity is $\sqrt{a^2+b^2}$. –  André Nicolas Aug 27 '11 at 23:20
    
Why cant I just do Velocity(final) = Velocity(initial) + at –  John Aug 27 '11 at 23:35
    
so 35m/s -10*2.5? –  John Aug 27 '11 at 23:36
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2 Answers

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For the magnitude of the velocity, it is $|v|=\sqrt{v_x^2+v_y^2}$, where $v_x$ and $v_y$ are the velocities in the $x$ and $y$ directions. For the direction part you should describe it as degrees above or below the horizontal. It will be $\arctan \frac{v_y}{v_x}$. As Rober Wiberg comments, you need to account for gravity changing the original velocity.

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Can you please The direction part in more detail, im greatly confused I dont know what arctan is. How did you get there? Where does this approach originate from. Thanks a lot. –  John Aug 27 '11 at 23:34
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If you imagine a right triangle with horizontal leg $v_x$, vertical leg $v_y$, the hypotenuse is $v$. The tangent of the angle with the horizontal is defined as $\frac{v_y}{v_x}$, and the arctangent is the inverse function to that. –  Ross Millikan Aug 27 '11 at 23:41
    
So then it doesent even have to be tan? cant i just the inverse of sin or cos? OR is there a special reason why I cant involve the hypotenuse, namely V ? –  John Aug 27 '11 at 23:46
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@John The 'inverse of cos' is arccos, the inverse of tan is arctan. You can form the whole triangle, so you can use either. –  mixedmath Aug 27 '11 at 23:58
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The key idea here is to separate the question into x and y components. So the overall velocity at $2.5$ seconds is not $\cos(24) \cdot 35$ - this is the initial x component velocity (although this does not change). The y component does change, due to the force of gravity. Use Newton to find that.

I see now that Ross answered with respect to the process of finding the magnitude and direction of the velocity. So I won't comment on that other than to emphasize the need to think component-wise.

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I have no problem with applying gravity, I just took a completely different approach to finding the direction. Can I take the magnitude of the velocity as a whole, then as individual components and form a triangle with all three. THen can I solve for the angle by using the cos formula? –  John Aug 27 '11 at 23:40
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@John: There is no good way to consider the effect of gravity on the velocity as a whole. It's effect is only clear on an independent y component. Then you can form a triangle with the x and y components to find the angle. –  mixedmath Aug 27 '11 at 23:47
    
alright I get that now thnx a lot :) –  John Aug 27 '11 at 23:50
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