An archer shoots an arrow with a velocity of 48.0 m/s at an angle of 46.5 ° with

Question

An archer shoots an arrow with a velocity of 48.0 m/s at an angle of 46.5° with the horizontal. An assistant standing on the level ground 150 m downrange from the launch point throws an apple straight up with the minimum initial speed necessary to meet the path of the arrow.

An archer shoots an arrow with a velocity of 48.0 m/s at an angle of 46.5A degree with the horizontal. An assistant standing on the level ground 150 m downrange from the launch point throws an apple straight up with the minimum initial speed necessary to meet the path of the arrow. What is the initial speed of the apple? Your response differs from the correct answer by more than 10%. Double check your calculations. m/s At what time after the arrow launch should the apple be thrown so that the arrow hits the apple? Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.

Explanation / Answer

x arrow = 48*cos(46.5)*t = 150

t = 150/(48*cos(46.5))=4.54 s

so y = v0y t - 1/2 g t^2 = 48*sin(46.5)*4.54 - 0.5*9.81*4.54^2=56.97 m

to reaach that height jus tbarely vf = 0

v^2 = v0^2 + 2 a y

v0 = sqrt(2*9.81*56.97)=33.43 s

b) it will take a time for apple to reach height v = v0 + at

t = 33.43/9.81 =3.41 s

so difference in times 4.54-3.41 =1.13 s

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