A projectile is shot from the ground at a velocity of #3 m/s# at an angle of #pi/3#. How long will it take for the projectile to land?

Answer 1

The time is #=1.06s#

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=3ms^-1#
The angle is #theta=(1/3pi)rad#
The acceleration due to gravity is #g=9.8ms^-2#
The distance #y=0#

Therefore,

#xtan(1/3pi)-(9.8*x^2)/(2*3^2cos^2(pi/3))=0#
#1.732x-1.089x^2=0#
#x(1.732-1.089x)=0#
#x=0#, this is the starting point
#x=1.732/1.089=1.59m#
The time is #t=x/v_x=1.59/(3cos(pi/3))=1.06s#
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Answer 2

The time of flight for the projectile is approximately 1.22 seconds.

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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.

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.

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.

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.

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