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A projectile is shot from the ground at an angle of #pi/6 # and a speed of #5 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Answer 1

Isabella Alvarez

#x=1,08 " "m#
#y=0,319" "m#

#y=v_i*t*sin alpha-1/2.g.t^2#
#y=5*0,25.0,5-1/2*9,81*(0,25)^2#
#y=0,625-0,306=0,319 #
#y=0,319" "m#

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Answer 2

Isabella Alvarez

The horizontal distance at the projectile's maximum height can be calculated using the formula:

[ \text{Horizontal distance} = \frac{{\text{Initial velocity}^2 \times \sin(2 \times \text{launch angle})}}{{\text{gravity}}} ]

Substitute the given values:

[ \text{Horizontal distance} = \frac{{5^2 \times \sin(2 \times \frac{\pi}{6})}}{{9.8}} ]

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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.