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A projectile is shot from the ground at an angle of #pi/4 # and a speed of #16 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=13,01 " meters"#
#y=6,52" meters"#

#t_m=(v_i*sin pi/4)/g#

#t_m=(16*0,707)/(9,81)#

#t_m=1,15" sec"#

#x=v_i*cos pi/4*t_m#

#x=16*1,15*0,707#

#x=13,01 " meters"#

#y=v_i*sin pi/4*t_m-1/2*g*t_m^2#

#y=16*0,707*1,15-1/2*9,81*(1,15)^2#

#y=13,01-6,49#

#y=6,52" meters"#

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Answer 2

Amelia Adams

The horizontal distance at the maximum height ((d_{\text{max}})) can be found using the formula: [d_{\text{max}} = \frac{\text{initial velocity}^2 \times \sin^2(\text{launch angle})}{2 \times \text{acceleration due to gravity}}] Substitute the given values: [d_{\text{max}} = \frac{16^2 \times \sin^2(\frac{\pi}{4})}{2 \times 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.