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A projectile is shot from the ground at an angle of #pi/8 # and a speed of #19 m/s#. When the projectile is at its maximum height, what will its distance from the starting point be?

Answer 1

Elena Albarran

The distance is #=13.02m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=19*sin(1/8pi)#

Acceleration is #a=-g#

At the maximum height, #v=0#

We apply the equation of motion

#v=u+at#

to calculate the time to reach the greatest height

#0=19sin(1/8pi)-g*t#

#t=19*1/g*sin(1/8pi)#

#=0.74s#

Resolving in the horizontal direction #rarr^+#

We apply the equation of motion

#s=u_x*t#

#=19cos(1/8pi)*0.74#

#=13.02m#

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Answer 2

Nicholas Agrusa

To find the distance at the projectile's maximum height, use the formula: [ \text{{Range}} = \frac{{\text{{Initial velocity}}^2 \times \sin(2 \times \text{{launch angle}})}}{{\text{{gravitational acceleration}}}} ]

Substitute values: [ \text{{Range}} = \frac{{19^2 \times \sin(2 \times \frac{{\pi}}{{8}})}}{{9.8}} ]

Calculate the result.

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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.