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If a projectile is shot at a velocity of #2 m/s# and an angle of #pi/6#, how far will the projectile travel before landing?

Answer 1

Emily Abbate

The distance is #=0.35m#

The equation describing the trajectory of the projectile in the #x-y# plane is

#y=xtantheta-(gx^2)/(2u^2cos^2theta)#

The initial velocity is #u=2ms^-1#

The angle is #theta=(1/6pi)rad#

The acceleration due to gravity is #g=9.8ms^-2#

The distance #y=0#

Therefore,

#xtan(1/6pi)-(9.8*x^2)/(2*2^2cos^2(1/6pi))=0#

#0.577x-0.0871x^2=0#

#x(0.577-1.633x)=0#

#x=0#, this is the starting point

#x=0.577/1.633=0.35m#

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Answer 2

Leo Abeyta

The horizontal distance traveled by a projectile can be calculated using the equation:

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

Where:

( v ) is the initial velocity (2 m/s)
( \theta ) is the launch angle (π/6 radians or 30 degrees)
( g ) is the acceleration due to gravity (9.8 m/s²)

Plugging in the values:

[ \text{Range} = \frac{{2^2 \cdot \sin(2 \cdot \pi/6)}}{9.8} ]

[ \text{Range} = \frac{{4 \cdot \sin(\pi/3)}}{9.8} ]

[ \text{Range} = \frac{{4 \cdot \sqrt{3}/2}}{9.8} ]

[ \text{Range} = \frac{{2 \cdot \sqrt{3}}}{9.8} ]

[ \text{Range} ≈ 0.36 , \text{m} ]

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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.