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December 17, 2014

December 17, 2014

Posted by **Sammie** on Monday, February 18, 2013 at 12:49am.

- Physics help!! -
**Elena**, Monday, February 18, 2013 at 4:28pmThe total range of the arrow is

L=vₒ²•sin2α/g =45²•sin100°/9.8 =203.5 m

R=150 m > L/2=101.75 m

=> the arrow is at its descending path

The time of the motion to the point of 'arrow-apple' meeting is

t₁=R/v(x)=R/v₀(x)= R/ v₀•cos α =150/45•cos50°=5.2 s

The time of the motion to the highest point is

t₀= vₒ•sinα/g=45•sin50°/9.8 =3.5 s

The time of the motion between the highest point and the point of meeting is

t= t₁ - t₀ =5.2 -3.5 = 1.7 s.

The maximum height of the arrow is

H= vₒ²•sin²α/2g =

= 45²•sin²50°/2•9.8=60.6 m/

The vertical downward displacement is

Δh=gt²/2=9.8•1.7²/2 =14.2 m

The position of the meeting point is

H =60.6-14.2 = 46.4 m (above the ground level)

Min. initial speed will be if the apple and the arrow meet at the apple’s top point

H=V₀t₂-gt₂²/2,

0= V₀-gt₂,

H=V₀²/2g =>

V₀=sqrt(2gH) = sqrt(2•9.8•46.4) =

=30.2 m/s

The time for this motion is

t₂=V₀/g = 30.2/9.8=3.1 s

Therefore, the apple has to start after Δt=t₁-t₂=5.2-3.1 =2.1 s.

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