An archer enters a particularly challenging archery contest. She issitting on a

Question

An archer enters a particularly challenging archery contest. She issitting on a boat that is floating down a river at a speed Vr=10m/s (assume the river is a straight line for this problem). Theboat is X=50 meters from shore, and there is a target right at theshoreline to aim at. However, the rule of the contest is that thearcher must fire the arrow when she reaches a point Y=20 metersupstream of the target, and the archer must fire directly at theshoreline (i.e. no firing forwards). Neglect air resistance forthis problem.

a) If the arrow is to hit the target, how much time must the arrowspend in the air?
b) What is the velocity of the arrow in the direction of theshoreline?
c) If the archer fires the arrow from a height 1m above the waterlevel, and the target is also 1m above the water level, what mustthe upwards component of the arrow's velocity be for the arrow tostrike the target?
d) At what angle (from the horizontal) is the archer going to haveto fire her arrow? What is the speed of the arrow when it hits thetarget?

Explanation / Answer

We take the direction of the river as the x-axis the directionto the shore as the y-axis and the up/down direction as thez-axis. She is moving at 10m/s in the x-direction the arrow startswith at least this speed in the x-direction. Furthermore she has toaim directly to shore (meaning that she can't add tothe x component of the velocity. >Conclusion: Since the arrow has to travel 20m in thex-direction (at 10m/s) to hit the target the arrow needs to be inthe air for 2 seconds. . Now since the shore is 50meters away and the arrow is in theair for 2 seconds we find that the velocity towards the shore(y-direction) is V=25m/s. . Since we are ignoring air resistance the motion in z-directionis a parabola. So she needs to give the arrow a z velocity suchthta at half way (t=1 second) the z-velocity is zero (i.e. it is atits highest point and ready to free fall the rest of theway, z=-1/2 g*t^2) Then the displacement in the z-directionis: .       (from t=0 to t=1where the z-velocityiszero)                   z=1/2*9.8*1^2=4.9m   .       We can calculate theinitial velocity in z (Viz) by using the following equationof motion: .                                                                                              Vfz2= Viz2 -2*g*z, then Vi=9.8m/s . In our configuration we have that (since she puts the initialvelocity of the arrow in the y-z direction): .                Viz=Vi*Sin   and Viy=Vi*Cos dividing these two we get Viz/Viy = Tan(since Tan=Sin/Cos) .               thensince Viy=25m/s and Viz=9.8m/s then Tan=0.392 then=arctan(0.392)=21.4 degrees. Notice that the target and the archer are at the same heightso at the point at the point it hits the target the velocity isequal to the initial velocity. So its speed is equal to themagnitude of the total velocity vector. We writte the vector: .            V=10m/si + 25m/s j + 9.8m/sk (where i,j,k are the usual unit vectors inthe x,y and z directions respectivelly. So then finally the speedat which the arrow hits the target is|V|=(102+252+9.82)=28.65m/s. . .

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