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A projectile is shot at an angle of #(5pi)/12 # and a velocity of # 17 m/s#. How far away will the projectile land?

Answer 1

Leo Abeyta

#14.45m#

Let us consider that the #"Velocity of projection"=u=17m/s and "Angle of projection"=alpha=5pi/12#

The vertical component of the velocity =#usinalpha#

The horizontal component of the velocity =#ucosalpha#

Let's additionally assume that the projectile's flight time is T.

The projectile must return to the ground within this T time, or until the net vertical displacement is zero after this "T" time.

in order for us to write

#0=usinalpha*T-1/2*g*T^2# #T=(2usinalpha)/g#

Horizontal displacement = distance from point of projection to the point of landing #="Horizotal component of velocity"xx T=ucosalphaxx(2usinalpha)/g=(u^2sin2alpha)/g=(17^2sin(2*5pi/12))/10m=(28.9*sin(pi-pi/6))=28.9*sin(pi/6)=28.9/2=14.45m#

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Answer 2

Charles Aeschliman

Use the range formula for projectile motion: ( R = \frac{v^2 \sin(2\theta)}{g} ). Given (v = 17 , \text{m/s}), (\theta = \frac{5\pi}{12}), and (g \approx 9.8 , \text{m/s}^2), calculate (R).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.