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If a projectile is shot at an angle of #pi/6# and at a velocity of #28 m/s#, when will it reach its maximum height??

Answer 1

Adrian Ayala

#10/7s#

#t =(usinalpha)/g=(28xxsin(pi/6))/9.8=(28xx1/2)/9.8=140/98=10/7s#

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Answer 2

Parker Abbott

The projectile will reach its maximum height at the midpoint of its flight when its vertical velocity component becomes zero. Use the kinematic equation: (v_f = v_i + at) and (v_f = 0), solve for (t) to find the time it takes to reach maximum height. Given that (v_i = 28 \sin(\pi/6) , \text{m/s}) and (a = -9.8 , \text{m/s}^2), (t = \frac{-v_i}{a}). Plug in the values to find (t).

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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.