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# Answer to Question #12166 in Mechanics | Relativity for zianna

Question #12166
A projectile is thrown with the speed of 25 meter per seconds in a direction 30° above the horizontal.
-how long does it take to reach the maximum height.
-how long does it reach the Projectile along in the air.
-how far does it go.
Expert's answer
Here is the formula for the height of a projectile:

H(t) = Vt&middot;sin&alpha; - gt&sup2;/2

Let&#039;s derivate it by t:

H&#039;(t) = V&middot;sin&alpha; - gt

V&middot;sin&alpha; - gt = 0 ==&gt; t = V&middot;sin&alpha;/g = 25&middot;sin30&deg;/9.8 &asymp; 1.2755 s.

-how long does it reach the Projectile along in the air.

The time of a flight is the time needed to reach the maximum height doubled:

T = 2t &asymp; 2&middot;1.2755 = 2.5510.

-how far does it go.

Using the obtained flight time:

L = VT&middot;cos&alpha; &asymp; 25*2.5510*cos30 = 55.2308 m.

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