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

Answer 1

To find the time it takes for the projectile to land, you can use the formula: ( t = \frac{2v_0 \sin(\theta)}{g} ), where ( v_0 ) is the initial velocity (2 m/s), ( \theta ) is the launch angle ((2π)/3), and ( g ) is the acceleration due to gravity (approximately 9.8 m/s^2). Plugging in the values, you'll get ( t \approx 0.408 ) seconds.

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Answer 2

Olivia Askew

#(2sqrt(3))/10# seconds

#T = (2u sin θ) / g#

#T = (2 × 2 ms^-1 sin ((2 π)/3)) / (10 ms^-2)#

#T = (4 × sqrt(3)/2) / (10)# s

#T = (2sqrt(3))/10# #text(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.