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A projectile is shot at an angle of #pi/4 # and a velocity of # 15 m/s#. How far away will the projectile land?

Answer 1

Isabella Alvarez

The range is #=23.0m#

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=15ms^-1#

The angle is #theta=(1/4pi)rad#

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

The distance #y=0#

Therefore,

#xtan(1/4pi)-(9.8*x^2)/(2*15^2cos^2(pi/4))=0#

#x-0.044x^2=0#

#x(1-0.044x)=0#

#x=0#, this is the starting point

#x=1/0.044=23.0m#

graph{x-0.044x^2 [-0.01, 14.04, -1.347, 5.673]}

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Answer 2

Elena Albarran

The range of the projectile is approximately ( \frac{15^2 \times \sin(2 \times \frac{\pi}{4})}{9.8} ) meters. Calculating this gives a range of about 11.47 meters.

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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.