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A projectile is shot at an angle of #pi/8 # and a velocity of # 1 m/s#. How far away will the projectile land?

Answer 1

Lily Adams

#d = 1^2/g * sin 45 = 1/(sqrt 2g) = 1/(10sqrt2) = sqrt 2/20 m#

We shall use the range formula

#v = 1m/s , theta = pi/8 =>2theta = pi/4 = 45^@#

#color(magenta)("Interesting note " 45^@ "is the optimal angle of launch")#

#d = 1^2/g * sin 45 = 1/(sqrt 2g) = 1/(10sqrt2) = sqrt 2/20 m#

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Answer 2

Sebastian Acosta

The horizontal distance ( D ) that a projectile will travel can be calculated using the equation:

[ D = \frac{{v^2 \cdot \sin(2\theta)}}{g} ]

Where:

( v ) is the initial velocity of the projectile,
( \theta ) is the angle of launch,
( g ) is the acceleration due to gravity (approximately ( 9.8 , \text{m/s}^2 )).

Substituting the given values:

( v = 1 , \text{m/s} ),
( \theta = \frac{\pi}{8} ),
( g = 9.8 , \text{m/s}^2 ),

[ D = \frac{{1^2 \cdot \sin(2 \cdot \frac{\pi}{8})}}{9.8} ] [ D = \frac{{1 \cdot \sin(\frac{\pi}{4})}}{9.8} ] [ D = \frac{{1 \cdot \frac{\sqrt{2}}{2}}}{9.8} ] [ D = \frac{\sqrt{2}}{2 \cdot 9.8} ] [ D = \frac{\sqrt{2}}{19.6} \approx 0.102 , \text{m} ]

Therefore, the projectile will land approximately 0.102 meters away.

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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.