Physics help!!

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An archer shoots an arrow with a velocity of 45 m/s at an angle of 50 degrees with the horizontal.An assistant standing on the level ground 150m downrange from the launch point throws an apple straight up with the minimum intial speed necessary to meet the path of the arrow. (a) What is the initial speed of the apple? (b) At what time after the arrow launch should the apple be thrown so that the arrow hits the apple?

• Physics help!! -

The 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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