Projectile Motion

Algebra Level 3

A cannon ball launched into the air had an initial velocity of 23.9 m/s. If the time of flight is 4.9s and the range is 10.05m, what is theta?

81 degrees 5 minutes 87 degrees 5 minutes 85 degrees 5 minutes 83 degrees 5 minutes

Erico Suňga by Erico Suňga , 27, Philippines

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2 solutions

Alex Li
Mar 13, 2015

Because the time is 4.9 s 4.9 s and the range is 10.05 m 10.05 m , the horizontal velocity is 10.05 4.9 = 2.051 m / s \frac{10.05}{4.9}=2.051 m/s . Now, the vertical velocity is 23. 9 2 − 2.05 1 2 = 23.81 m / s \sqrt{23.9^2-2.051^2}=23.81 m/s , so the angle is arctan ⁡ ( 23.81 2.051 ) ≈ 8 5 ∘ 5 ′ \arctan(\frac{23.81}{2.051})\approx\boxed{85^\circ 5'}

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Erico Suňga
Aug 5, 2014

http://sharic091910.wordpress.com/2014/08/06/solution/

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