A projectile is shot from the ground at a velocity of #6 m/s# and at an angle of #(7pi)/12#. How long will it take for the projectile to land?
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To find the time of flight, we can use the horizontal component of the initial velocity (V_{i_x}) and the horizontal displacement. (V_{i_x} = V_i \cdot \cos(\theta)) (V_i = 6 , m/s) (given) (\theta = \frac{7\pi}{12}) (given)
(V_{i_x} = 6 \cdot \cos\left(\frac{7\pi}{12}\right)) (V_{i_x} \approx -2.12 , m/s) (approximate value)
Since the projectile lands back at the same height from where it was launched, the vertical displacement (y) is zero. We can use the vertical motion equation: (y = V_{i_y} \cdot t + \frac{1}{2} a t^2) (V_{i_y} = V_i \cdot \sin(\theta)) (V_{i_y} = 6 \cdot \sin\left(\frac{7\pi}{12}\right)) (V_{i_y} \approx 3.54 , m/s) (approximate value)
Since the projectile lands back at the same height from where it was launched, the vertical displacement (y) is zero.
(0 = (3.54) \cdot t + \frac{1}{2} (-9.8) \cdot t^2)
Solving this quadratic equation will give the time of flight, (t). (t \approx 0.717 , s) (approximate value)
Therefore, it will take approximately 0.717 seconds for the projectile to land.
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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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