Question

Problem 4.

An archer shoots an arrow with a velocity of 30 m/s at an angle of 20 degrees with respect to the horizontal. An assistant standing on the level ground 30 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 and at what time after the arrow is launched should the apple be thrown so that the arrow hits the​

Answer 1

Explanation:

The motion of the arrow is a projectile motion. Then the motion of the arrow along horizontal direction is a motion with constant velocity:

x(t)=VxT

Where the initial position is 0 and Vx=30*cos(20degree)=28.2m/s . Then

x(t)=28.2t

Motion along vertical axis is a motion with constant acceleration, then

y(t)=VyT-gt^2/2

Where g=9.8m/s^2 , initial position is 0, and vy=30*sin(20degree)=10.3 m/s . Then

y(t)=10.3t-4.9t^2

Then we need to find the time when the arrow will be exactly above the assistant. At this moment of time x(t) =100. Then we can find the time:

t=Tarrow=30/28.2=1.06s

Then we can find the height of the arrow at this moment of time:

y(t=1.06)=1.06*1.03-4.3*1.06^2=5.4m

The assistant should through the apple with minimal velocity so it will reach point 5.4 m and at this height the velocity should be 0. From this condition we can find the minimal velocity:=10.3m/s

The time of the motion of the apple to this point is=1.05s

Then the apple should be thrown after=.01s