A projectile is shot from the ground at a velocity of #8 m/s# and at an angle of #(2pi)/3#. How long will it take for the projectile to land?

Answer 1

To calculate the time of flight for the projectile, use the formula: ( \text{time} = \frac{2 \cdot \text{initial velocity} \cdot \sin(\text{launch angle})}{\text{acceleration due to gravity}} ). Substituting the given values: ( \text{time} = \frac{2 \cdot 8 \cdot \sin(\frac{2\pi}{3})}{9.8} ). Calculate to find the time.

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Answer 2
#|vecv| = 8m*s^(-1)#
#vecv_x = 8cos((2pi)/3) m*s^(-1) hati=-4 hati#
#vecv_y = 8sin((2pi)/3) m*s^(-1) hatj=4sqrt3hatj#

On the y axis,

#v_i = 4sqrt3#

When standing at your highest point,

#v_f=0#
#a=-10 m*s^(-2)#

Using the initial equation of motion,

#0=4sqrt3 - 10t rArr t=0.4*sqrt3 rArr t=0.692s#

Thus, it takes 0.692 seconds to reach the maximum height, and it takes 0.692 seconds for the projectile to descend to the ground from the highest point.

#"Total time taken" = 2*0.692 = 1.384s#
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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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